P5.7 Solving trig equations

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was designed to test candidates use of trigonometric identities and solving trigonometric equations and most candidates attempted this question, with a significant number achieving at least one mark. Only a small minority of candidates did not attempt this question. Most candidates were able to recognise the use of one of the three trigonometric identities. Errors were generally in substitution where some candidates used x instead of 3x. Where candidates correctly substituted the trigonometric identity correctly they usually rearranged to the correct format, thus gaining the first two marks. A common error from this point was to miss out the square rooting and use the arcsin/arctan/arccos with the constant, thus scoring no further marks. A small number of candidates square rooted, however did then not proceed to the required format by missing out the division by 3, scoring no more than two marks. For candidates who correctly square rooted and then solved for 3x, a large majority did not achieve the final A mark as they did not consider both roots and/or change the scale for the question. It was common for candidates to achieve one or two solutions, rather than all three. Only a very small number of candidates found, additional wrong solutions within the range provided. A few gave additional solutions outside the range although this was not penalised.

Common mistakes

• use sin²3x/cos²3x = tan²3x (or sin² + cos² = 1): tan²3x = 4
• take the square root BEFORE inverse tan, then divide by 3
• x = 21.1°, 38.9°, 81.1° and no others in 0–90°
• using x instead of 3x
• forgetting to divide by 3

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was a good differentiator at the later end of the paper. Success with this question was somewhat mixed, with some learners scoring most marks in part (b), whilst part (a) proved to be more challenging as it tested not only their knowledge of trigonometric identities but also their algebraic manipulation skills. Part (b) was generally answered well, most responses used the result in part (a) to set up the equation 2cos 3sin 2 = theta theta and then proceeded to reach and equation in cos and sin theta theta by applying the double angle formula on the right hand side, gaining the first method mark. From here, almost all responses reached the equation 1 sin 3 = theta , through either factorising correctly ( ) ( ) ( ) i.e. 2cos 1 3sin or cos 2 6sin - - theta theta theta theta , or dividing both sides by cos . theta Those that divided both sides by cos theta usually did not find that there was a further solution, 90 . = ° theta Those learners who reached 1 sin 3 = theta or equivalent, were usually able to gain the dependent method mark by finding at least one of the solutions of the equation in the given range. Responses that factorised the equation correctly, usually gained full marks by finding all solutions. Learners who used a calculator solely to find the correct angles scored no marks. 21

Common mistakes

• use (a): 2cos θ = 3sin 2θ = 6sin θ cos θ
• factorise cos θ(1 − 3sin θ) = 0 to get θ = 90° as well as sin θ = 1/3 giving 19.5°, 160.5°
• dividing by cos θ loses 90°
• calculator-only solutions score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question, which required an understanding of transformation and periodicity of trigonometric functions, gave learners the opportunity to think carefully and demonstrate a strong understanding of the topic. Part (i) was generally well answered with many responses evidencing use of graphs to help 1 deduce that n was 3. The most common incorrect answer was , even though the n = 3 question specified that n was a natural number. Learners would be well advised to pay careful attention to the constraints placed on unknowns in all questions. In part (ii), a good number of responses demonstrated that there were 30 solutions to the equation, although fewer were able to communicate their reasoning well enough. A significant minority arrived incorrectly at 30 by mistakenly assuming that there were 15 solutions to sin(nx) = k and 15 solutions to sin(nx) = –k, or sometimes 18 and 12 solutions respectively (the correct number of solutions being 16 and 14 respectively). Such responses scored one of the two available marks. An uncommon but elegant approach was to rewrite the quadratic in sin3x as a linear equation in cos6x, making it easier to explain why there are 30 solutions in the given interval. Another was to identify 2 3 as the period of sin2 3x and then scale up accordingly, which demonstrated an excellent understanding of periodicity. A small number did not realise that n was still equal to 3 in part (ii), and gave an answer in terms of n. A common incorrect approach involved the misapprehension that squaring sinx would also square the number of solutions, leading to an incorrect answer of 36. Many did not take the negative solutions into account and simplify multiplied 6 by 2.5 to obtain the incorrect answer of 15. More learners would have benefited from a well-drawn and labelled graph to support their thinking and reasoning.

Common mistakes

• sin²(3x) = k² means sin 3x = ±k: 6 solutions per interval of length π, 5 intervals gives 30
• justification needed (16 + 14, not 15 + 15 or 18 + 12)
• 36 and 15 are common wrong answers
• a labelled sketch helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The trigonometric equation, again, demonstrated how candidates are usually successful in solving these types of equations, however, they often struggle with the manipulation to show a given result. Part (b) was much more accessible due to the given answer in (a) not being a “show” which required constants to be found. Candidates achieved much more success with this part with most able to find the required roots and use these to find at least two of the required angles. There were occasional sign errors and a very small number attempted to solve the original equation from (a) without success. There were quite a few candidates who did not find all the required angles. This would often be because they would reject one of the solutions, 2 3 - or the angle it gave being negative seemed to be a reason that candidates ruled it out. The other root, 2, would often be ruled out presumably because of a misconception that the range of all trigonometric graphs was |𝑦| ⩽1. There were a small number who used an incorrect method to find other values, such as adding 90, or subtracting from 180 from their angles. Quite a lot of candidates did manage to find all the required angles, however.

Common mistakes

• roots tanθ = 2 and −2/3
• four answers 63.4°, 146.3°, 243.4°, 326.3°
• rejecting the negative root or missing the third/fourth quadrant solutions

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The trigonometric equation, again, demonstrated how candidates are usually successful in solving these types of equations, however, they often struggle with the manipulation to show a given result. Part (c) was usually poorly attempted with the vast majority not evidencing an understanding of graph transformations and just adding 63.4 to 720 or multiplying 243.4 by 4 as this would give an angle in the required range.

Common mistakes

• 4α = θ + 360n: 4α in (2880°, 4320°) gives 4α = 2880 + 63.4, α = 735.9°
• common errors adding 63.4 to 720 or 4 × 243.4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This harmonic form question was seen to be relatively challenging, despite similar questions being seen on several previous exam papers. That said, many candidates were able to attempt all parts of the question, and many follow through marks were available to those that made errors early on in their attempt. It was pleasing that most candidates identified that this question used degrees rather than radians. It was also pleasing to see many candidates continue to work through all parts of this modelling question despite having errors or not attempting previous parts. Part (d) was accessible to most candidates and fully correct answers were not uncommon. The simplest and most successful approach was to set ( ) sin 30 70 1 t + = - and solve for t, then substitute this into their model for R. Some candidates incorrectly set ( ) sin 30 70 1 t + = or 0, leading to the equation 30 70 90 t + = or 180. To score full marks for this part of the question, candidates needed to find a value of t from 30 70 270 t + = , use it with sufficient accuracy when substituting into their expression for R and round their final answer to an integer. Some candidates obtained a correct decimal answer but then rounded to 3 significant figures, giving 1030 as their final answer rather than the required 1032 or 1033. Many candidates incorrectly found a value of 16 3 t = - (coming from 30 70 90 t + = - ) which should have been an indication to restart this part having recognised that time cannot be negative. A significant proportion of candidates used 90 on the right-hand side, or even 0 or 180, which were not acceptable, and these failed to score any marks. There was also a minority of candidates who thought they were being asked to find the value of F. Another minority of candidates incorrectly assumed the question was asking when the populations of foxes and rabbits were the same and wrote down equations which they quickly concluded they could not solve.

Common mistakes

• fox minimum when sin(30t + 70)° = −1 so 30t + 70 = 270 (not 90 or 180), t = 20/3
• substitute into the rabbit model
• whole number 1032

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A significant number of candidates found this question challenging. In part (b), candidates that spotted the connection with part (a) tended to achieve at least 3 out of the 4 marks available. Those who failed to get the first line right with the variable of 2x on both sides but just substituted 2tan sec x x or even 2tan sec on the left-hand side gained no marks in this part. Candidates took a number of different approaches to solving the correct equation. The most frequent was to obtain 2 1 tan 2 2 x = but approaches involving 2 3sin 2 1 x = or 2 3cos 2 2 x = were relatively common. There were a few candidates who used the identity 2 2tan tan 2 1 tan x x x = - to reach a quadratic in 2 tan .x These attempts were usually successful. Some candidates stated that the square root of 1 2 is 1 4 which lost them the final two marks in this part. However, those candidates that did get to one of the correct equations could normally proceed to one correct answer (which was typically x = 17.6°). It was only a minority of candidates who gave both correct answers. The final mark was regularly lost because only one solution was identified, or candidates had incorrectly found 45 degrees by incorrectly solving sec2x = 0 and interpreting this as cos2x = 0 rather than being undefined. Most candidates that achieved marks in this part of the questions could correctly identify the order of operations to solve their equations although there were some candidates who struggled with this. For the final mark, a few wrote down 17.7 instead of 17.6 and some had extra answers in the given range and therefore lost the accuracy mark. A small minority of candidates did not show appropriate working in part (b), sometimes merely writing down the two correct answers to the given equation. The question explicitly asked for all stages of a candidate’s working as the answers can be found using a calculator. Hence candidates that gave the correct answers without appropriate working were not awarded any marks in this part of the question.

Common mistakes

• replace θ by 2x: 2tan2x sec2x = cot2x sec2x
• sec2x ≠ 0 (not 45° from cos2x = 0)
• tan²2x = ½ (or sin²2x = ⅓)
• x = 17.6°, 72.4° (both, 1 d.p., none extra)
• answers without working score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question involving the use of trigonometric identities and solving a trigonometric equation was generally very well answered with the exception of part (c). There were very few blank scripts. Part (b) was attempted by the majority of candidates with some good success. Calculators will solve this equation directly, and as a result there were some candidates (although a minority) who were unable to work out the correct answer using a clear written method combining the quadratic formula and using the inverse candidates were able to solve the equation and find at least one value of x (usually awrt 42.6). Generally, where candidates understood the requirements of the question, both angles were found to the required accuracy. Some candidates stated additional incorrect answers inside the given range and so lost the final accuracy mark. It was surprisingly common to see candidates rounding their values from the quadratic equation (i.e., to 0.7) which resulted in inaccurate angles and as a result they lost the final accuracy mark.

Common mistakes

• sin x = (−2 + √29)/5 ≈ 0.677 (reject the root below −1)
• x = 42.6° and 137.4°
• rounding sin x to 0.7 loses accuracy
• extra solutions in the range lose the final mark
• show the quadratic being solved

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question involving the use of trigonometric identities and solving a trigonometric equation was generally very well answered with the exception of part (c). There were very few blank scripts. Part (c) was much less successful, with only a small number of candidates gaining any marks. Most candidates only identified one of the adaptations to the question: either the stretch or the increase in the range. The most common incorrect answer resulted from them multiplying their number of solutions by 5 to achieve a total of 10 solutions. Of those candidates who realised there were 30 solutions nearly all were able to give a satisfactory explanation.

Common mistakes

• 3x stretches: 3 periods per 360° and 1800° is 5 lots of 360°, so 15 × 2 = 30 solutions
• multiplying by 5 only (10) is the common error
• explanation required

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b) was generally done well by the candidates who managed to obtain a quadratic in part (a), with many candidates correctly solving to get 360 or 540, and then solving their quadratic. Those who found a solution to their quadratic also tended to find solutions based on that within the specified set of values. Where marks were lost, it was often by incorrectly including extra values,in particular the value of 450 degrees which had been specifically excluded in the question.

Common mistakes

• sin2x = 0 gives 360° and 540° (450° is excluded: tan x undefined)
• the quadratic factorises as (23cos x + 15)(cos x − 1), cos x = −15/23 gives 491°
• answers 360°, 491°, 540° only, including 450° is the common error

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

the final answer was stated exactly as printed in order to achieve all the marks. In part (c) most candidates realised they should solve the given quadratic in tan alpha to give two possible angles of projection. Consideration of the situation should have indicated that the larger of the two angles would lead to the greatest possible height. Although many recognised that the greatest height is reached when the vertical component of velocity is zero, some chose to use the smaller 45o angle. Some found the greatest heights for both angles but then failed to select the relevant one. Other errors included using the time to greatest height as half the answer in part (a) or choosing sin alpha = 1 to give the maximum height in

Common mistakes

• tan α = 1 or 3: the larger angle gives the greatest height
• vertical velocity zero: 0 = (28 sin α)² − 2gH, H = 36 m
• not using half the time from (a) or sin α = 1

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

There were a number of completely blank responses to both parts of this question. This was either due to timing on the paper or possibly due to this topic not having been covered in sufficient depth due to disruption to learning. In part (b) it was pleasing to see candidates making an attempt although there were quite a few with nothing at all correct. These often had incorrect or irrelevant identities, resulting in equations which could not be solved for sin 2x, meaning that they could not access the last method mark. Those who did achieve 2 sin 2 3 x = (or -1) usually solved correctly to get, at least, one value of x, usually x = 20.9 and very often the second value too. Most who got to 11 this stage usually rejected any other values and scored full marks. Often, they had cancelled the cos 2x earlier in their working, so they did not have to reject the extra values at the end. Usually this should be avoided, although on this occasion the domain meant that this did not affect their final answer. It was pleasing to see that candidates had not tried to answer the question entirely by calculator by just stating the angles and the use of radians rather than degrees was not typically seen. Many did not recognise the relevance of part (a) and began this part from scratch, often successfully, even though they had not shown any relevant working for part (a).

Common mistakes

• multiply by cos 2x and use cos²2x = 1 − sin²2x (or use (a) with θ = 2x): 3sin²2x + sin 2x − 2 = 0
• sin 2x = 2/3 (reject −1 and x = 45° as cos 2x ≠ 0)
• find 2x then halve: 20.9°, 69.1°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was another big discriminator. Good candidates found it familiar and accessible with some 25% scoring full marks. Strangely, for candidates who dropped marks, they seemed to be able to do either part (a) or part (b), but not both. Part (b) could be attempted without having done part (a) and many took up this opportunity. Those who paid attention to the 'hence', deducing that x = 2 theta + 60, were very successful in achieving the correct answers, though quite a few only found one solution. For the many students who did not see the link between parts (a) and (b), generally encountered much more complex, longer algebraic manipulation. Such students re-started the question using the compound angle formulae with a combination of other trigonometric identities, but were often not able to reach an equation in one function, or else made numerous slips. If quadratics in tan 𝜃𝜃 were reached, they were dealt with competently. If the candidate’s efforts led to tan2 theta = k, k≠ 3 sqrt 3 or tan (2 theta ± 60) = 3 sqrt 3 they were generally able to use the correct order of operations to solve their equation to find at least one solution. A fair number of correct answers were seen written down without any correct working. As the question had a warning about the use of calculator technology, these responses gained no marks

Common mistakes

• x = 2θ + 60°: tan(2θ + 60°) = 3√3 gives 2θ + 60° = 79.1°, 259.1°
• θ = 9.6°, 99.6° (second solution often missed)
• restarting without the link is much longer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was generally well done with many candidates scoring 3 out of 4 but very few scoring 4 out of 4. Candidates generally understood that the question required an initial differentiation and this was done correctly on the whole. However, incorrect attempts at differentiation included: • 16cosቀ 1 2 𝑥𝑥ቁ-3 • 8cosቀ 1 2 𝑥𝑥ቁ-3 • -4cos ቀ 1 2 𝑥𝑥ቁ-3 Candidates also understood that they had to set their derivative to zero which required a combination of manipulating the algebra, applying arccos, and doubling their answer. The difficulty encountered in this part was the recognition that it was the 3rd solution which was required and most gave either the 1st or 2nd turning point. As a diagram was given in the question, candidates should use this to help them understand what is required. Some students who gave the 1st turning point also checked by further differentiation that their point was a maximum. A few candidates unnecessarily found the corresponding y value for their x value. In establishing their value of x, it was disappointing to see all of the following: • candidates halving instead of doubling • using degrees • arriving at a correct answer but leaving it as 14.01 instead of 14.0

Common mistakes

• f′(x) = 4cos(½x) − 3 (not 8cos or 16cos)
• cos(½x) = ¾, double (not halve) the angle
• use the sketch to pick the third turning point x = 14.0 (not 1.45 or 11.1)
• radians
• 14.0 not 14 or 14.01

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The majority of candidates found this question accessible and were able to score consistently throughout. Where marks were lost they were usually in parts (b) and (c). 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 6𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎 Part (a) was attempted successfully by most candidates, starting usually with 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. Candidates then successfully cross multiplied and used the was 6𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡= 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቀ 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቁ 2 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 appropriate identity to reach the required result. Another starting point regularly seen . Errors in the proof were rare if candidates started from a correct statement. Occasionally the “ = 0” was carelessly missing. 𝜋𝜋 In part (b) most candidates correctly solved the equation either by factorising or using 6 rather than the required 5𝜋𝜋 5𝜋𝜋 calculator technology, but many reached the acute angle 6 . Others found a number of solutions but failed to identify 6 as the obtuse angle of 𝜋𝜋 required answer. A few candidates worked in degrees and again lost the accuracy mark. 6 in (b) then usually carried this error into (c) and 𝜋𝜋 Those who had obtained an angle of 6 in (c) often would reach 9൫1 + sqrt 3൯, but did not seem to worry that this did not match the required answer were in a position only to earn the method marks. Candidates using format given in the question. There were many attempts, however, to use the correct formula for the sum to infinity with appropriate values for the first term and the (often incorrectly) algebraic expressions for the sum to infinity involving 𝜃𝜃, without common ratio. Some candidates unfortunately put effort into finding and rearranging 𝑎𝑎 1+𝑟𝑟 incorrectly as the formula substituting an angle in, gaining no credit, and others used for the sum to infinity. The most common mistake in (c) was failure to show the steps 3 ൰ straight into their calculators to obtain 9 -9 sqrt 3 losing the last two marks in -6 sqrt 3 where the denominator was rationalised. Many candidates clearly typed an expression 1-൬- sqrt 3 like the question.

Common mistakes

• sinθ = ½ (25/2 impossible)
• θ obtuse so 5π/6, not π/6 or 150°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates were usually able to make good progress on the first half of this question on solving trigonometric equations, but they tended to run into problems halfway through. It was rare for full marks to be scored, although most were able to score around half of the available marks. In part (i), nearly all candidates were able to proceed to the correct quadratic in sin , although the occasional candidate tried to incorrectly substitute sin into the right- = - 1 cos hand side of the equation. Most candidates were able to apply the quadratic formula and proceed to finding all three angles. Some found additional angles, which even if they are outside the range, prevent full marks from being scored. Most candidates seemed to opt for sketching trigonometric graphs to find the other angles as opposed to using the CAST diagram. Part (ii)(a) was poorly answered, however. Most candidates failed to score either mark, although those who did were usually able to make some reference to cancelling out sin x . Just stating that not all the angles had been found was insufficient explanation as to the errors made so candidates should be encouraged to provide greater detail if they are able to do so. Part (ii)(b) was attempted by most candidates, with many able to equate to the correct angle, rearrange and proceed to the correct answer. There were a number who equated to several angles, but either did not make their final answer clear, or did not rearrange to find the required angle. 8

Common mistakes

• use cos²θ = 1 − sin²θ to get 5sin²θ + 6sin θ − 5 = 0
• sin θ = (−3 + √34)/5 only (other root < −1)
• θ = 34.5°, 145.5°, 394.5° in 0 < θ ≤ 450°
• extra values outside the range lose the final mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates were usually able to make good progress on the first half of this question on solving trigonometric equations, but they tended to run into problems halfway through. It was rare for full marks to be scored, although most were able to score around half of the available marks. In part (i), nearly all candidates were able to proceed to the correct quadratic in sin , although the occasional candidate tried to incorrectly substitute sin into the right- = - 1 cos hand side of the equation. Most candidates were able to apply the quadratic formula and proceed to finding all three angles. Some found additional angles, which even if they are outside the range, prevent full marks from being scored. Most candidates seemed to opt for sketching trigonometric graphs to find the other angles as opposed to using the CAST diagram. Part (ii)(a) was poorly answered, however. Most candidates failed to score either mark, although those who did were usually able to make some reference to cancelling out sin x . Just stating that not all the angles had been found was insufficient explanation as to the errors made so candidates should be encouraged to provide greater detail if they are able to do so. Part (ii)(b) was attempted by most candidates, with many able to equate to the correct angle, rearrange and proceed to the correct answer. There were a number who equated to several angles, but either did not make their final answer clear, or did not rearrange to find the required angle. 8

Common mistakes

• dividing by sin x loses sin x = 0, so x = 0 is missed
• also missed x = −53.1° from cos x = 3/5 in the range
• "not all angles found" alone is not enough

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates were usually able to make good progress on the first half of this question on solving trigonometric equations, but they tended to run into problems halfway through. It was rare for full marks to be scored, although most were able to score around half of the available marks. In part (i), nearly all candidates were able to proceed to the correct quadratic in sin , although the occasional candidate tried to incorrectly substitute sin into the right- = - 1 cos hand side of the equation. Most candidates were able to apply the quadratic formula and proceed to finding all three angles. Some found additional angles, which even if they are outside the range, prevent full marks from being scored. Most candidates seemed to opt for sketching trigonometric graphs to find the other angles as opposed to using the CAST diagram. Part (ii)(a) was poorly answered, however. Most candidates failed to score either mark, although those who did were usually able to make some reference to cancelling out sin x . Just stating that not all the angles had been found was insufficient explanation as to the errors made so candidates should be encouraged to provide greater detail if they are able to do so. Part (ii)(b) was attempted by most candidates, with many able to equate to the correct angle, rearrange and proceed to the correct answer. There were a number who equated to several angles, but either did not make their final answer clear, or did not rearrange to find the required angle. 8

Common mistakes

• 5α + 40° = 720° − 53.1° for the fourth positive solution, giving α = 125°
• list the solutions in order and rearrange for α
• make the final answer clear

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was found to be another discriminating question. Those who were familiar with the topic made good progress yet there were many other attempts that scored no marks at all. Part (b) A significant number of candidates used the result of (a) correctly and started their solution with tan2x=3sin2x. However, it was also the case that tanx =3 sin2x was seen several times as well. Of those candidates who started correctly with tan2x = 3sin2x, the significant majority simply cancelled out the resulting factor of sin2x and in so doing neglected to consider the possibility that sin2x=0 and in so doing the solution x = 90° was omitted from the fully correct list of three answers. Many candidates however, did state that cos2x =⅓ and went on to get 35.3° and 144.7° and so only one mark was lost.

Common mistakes

• replace θ by 2x: tan 2x = 3 sin 2x (not tan x)
• factorise sin 2x(1 − 3cos 2x) = 0, do not cancel sin 2x (loses x = 90°)
• x = 35.3°, 90°, 144.7°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between candidates that got this far, but it did appear that many candidates had simply run out of time, perhaps indicative of a generally lower understanding than in a June series because of it being a resit paper following the pandemic. Part (c) was another challenging part to this question and many candidates made no attempt. A surprising proportion of candidates attempted to solve the equation cos(0.5𝑡𝑡+ 𝛼𝛼) = -1, apparently in an attempt to find the minimum, but did not appear to understand how to use this to solve the problem. It was uncommon to see this method result in a correct value for 𝑇𝑇. More commonly, candidates were successful in finding a value for 𝑡𝑡 that satisfied 𝐻𝐻= 0, but incorrectly interpreted this as the solution to the problem. Instead, they were required to find a second value of 𝑡𝑡 and find the difference between these times. Again, there were a number of efficient and concise solutions to the problem that demonstrated an excellent understanding of what was required.

Common mistakes

• set H = 0: cos(0.5t + 0.464) = −3/(2√5)
• need two consecutive solutions and their difference (one time is not T)
• 3.34 exactly to 3 s.f.

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was attempted by most candidates with many being able to secure nearly all the marks in part (c). It was rare, however, for full marks to be scored overall, with most struggling with the transformations in part (b). Part (c) proved to be high scoring for many candidates, however, some did not recall the identities correctly and made little progress. In particular, candidates made errors with rearranging 2 2 sin cos 1 x x + = or missing the square from either or both terms. The given interval was unusual, although it was pleasing to see many of the candidates who proceeded to find a value for were able to find the one that was in the required interval. 9

Common mistakes

• multiply by cos θ: 3cos²θ = 8 sin θ, then cos²θ = 1 − sin²θ
• 3sin²θ + 8sin θ − 3 = 0 gives sin θ = ⅓
• only 520.5° lies in 450 to 720

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was the second modelling question on the paper. Part (c) was more challenging. The correct answer could be obtained fairly quickly by solving 1.107 3 12 2 t + - = pi pi but often the 1.107 and/or the 3 were omitted. Stating the time in an acceptable format also proved surprisingly difficult.

Common mistakes

• solve πt/12 + 1.107 − 3 = π/2 in radians, t = 13.2
• the 1.107 and the 3 are often omitted
• give a time of day, 13:14 (1:14 pm)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b), however was very often left completely blank. Many who attempted it failed to see the connection with (a) and got lost in complicated trig expansions which led nowhere. Of those who did use part (a) to answer part (b), the most common solution was simply 𝑥𝑥=25 after cos 𝑥𝑥 had been cancelled, thus losing the solution given by cos 𝑥𝑥 =0. Only a small minority of the most able found all 3 solutions.

Common mistakes

• use (a) to get cos x cot x = cos x cot(3x − 50°)
• do not cancel cos x without cos x = 0 (x = 90°)
• cot x = cot(3x − 50°) gives x = 25° and, as cot has period 180°, x = 115°
• three solutions

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b), it was common for candidates to achieve no marks for part (a) and then full marks for this part. Most of the students gained the first method mark and managed to produce an equation in cosx only. Mistakes were made with signs and brackets. In particular, candidates failed to write brackets around 4cos3x – 3cosx and hence would obtain an incorrect quadratic equation to solve. Several candidates ‘solved’ the cubic/quadratic equation using the polynomial equation function on their calculator and did not show any evidence of their working. Some incorrect equations led to complex solutions but most of these candidates still did not realise that they had made an error. A large number of candidates divided by cosx to simplify their expression but they failed to recognise this lost a solution and therefore were not awarded the final mark. A small number of candidates failed to gain any marks on this part as they replaced the right hand side with cosx but left the left hand side with cos3x and then proceeded to treat this as a cubic/quadratic. When candidates had identified the correct equation and solutions they were very good at finding all the solutions in the range. A large number who did get solutions did not include -90° or included 180° and therefore lost the final mark.

Common mistakes

• brackets round 4cos³x − 3cos x
• factor out cos x, do not divide by it (loses cos x = 0)
• solutions −90°, 0°, 90°, 139°
• include −90°, exclude 180°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many candidates left the entire question or parts of this question completely blank and clearly found it difficult to access the content being assessed here, perhaps due to a lack of familiarity with integration using parametric equations. Many tried to establish the Cartesian equation to answer various parts. That said, there were many candidates who provided clear and concise solutions to each part. In part (b) many candidates who had found part (a) difficult gave up and lost the opportunity of scoring some easier marks. Many candidates who scored no marks elsewhere were still able to access at least the first two marks in part (b) by setting sin2t = 4.2 5 and solving, frequently getting at least one correct angle; but too many of these found their second value of t by subtracting their first value from pi instead of calculating pi – 𝑠𝑠𝑠𝑠𝑠𝑠-1( 0.84) before dividing by 2. The following mark was for finding two values of x, but many simply substituted their first value in to find x, thinking that this gave them the width of the walkway. Too many marks were lost by approximating too soon and hence not having values with the required accuracy. A few lost the last mark because they omitted the units.

Common mistakes

• 5sin 2t = 4.2 gives two values of t: 2t = π − arcsin 0.84 for the second (not π − first t)
• find x at both and subtract
• width 2.40 m (units needed)
• avoid early rounding

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b) was, on the other hand, very badly done. Many candidates failed to heed the use of the word ''Hence'' and started again wasting valuable time. Of those who did use part (a) failed to have a reliable method of solving 4 5cos 4 3sin x x with may failing to eliminate the 4's. Of those who did reach 5cos 3sin x x many had solutions that involved variations of 3 tan 0 5 x . Better candidates found the manipulation straightforward and proceeded to find both solutions without any difficulty. As a result this question was an effective discriminator.

Common mistakes

• use (a): 4 − 5cos x = 4 + 3sin x so tan x = −5/3
• x = 121°, 301°
• avoid starting again

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was generally a well attempted question with most candidates able to score over half of the marks. In part (a) the correct identities were well known and efficiently used in order to obtain an equation in one function. Almost all candidates who reached this stage were then able to proceed to find at least one correct angle. It was clear from the graphs and diagrams sketched by most candidates that they had been taught to look for all angles in the required range and many were able to give all four of the angles needed. Only a minority of candidates identified 0, 180 as solutions to the equations. Most candidates divided through by sin or cos and failed to identify any of these values.

Common mistakes

• sin 2θ = 2 sin θ cos θ and tan θ = sin θ/cos θ
• 10cos²θ = 9 gives ±18.4°, ±161.6°
• dividing by sin θ loses 0° and ±180°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was generally a well attempted question with most candidates able to score over half of the marks. In part (b) a significant number used 2 50 x rather than 25 x .Of those who used one of their solutions to part (a), the majority then proceeded to the correct answer, although a significant number did not give the smallest positive solution to the equation. This was usually due to the fact that they used their smallest positive solution to part (a) (i.e. 18.4) in the equation, not realising that they needed to use – 18.4.

Common mistakes

• θ = x − 25 (not 2x − 50)
• use θ = −18.4° (not 18.4°) to get the smallest positive x = 6.6°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between students of all abilities with part (a) and (b) more accessible to students compared to part (c) and (d). There was less success in part (c). Attempts at this part were varied, with a sizeable proportion not making the connection with part (a) at all and attempting to find the value of H(t) at pi /2 or some other value, or failing to substitute into H(t) at all. Many solved tan x = 4 to find the acute angle but did not then look for the further solution which was needed for the height between the first and second bounces and so achieved only the first mark. Also, use of degrees instead of radians was a common error, but this was allowed the first method mark. The magnitude of their resulting answer should have alerted them to the fact that something had gone very wrong.

Common mistakes

• need the second solution of tan t = 4, t = 4.47 (not 1.33)
• radians not degrees
• H ≈ 3.18 m

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q12(a) discriminated well between the higher ability candidates and it was possible for candidates of all abilities to gain access to Q12(b). In Q12(a), many candidates struggled to apply a complete strategy which would help them to make significant progress in proceeding from cos3 sin3 , via an sin cos 2 2 cos2 (cos sin ) intermediate stage of cos(3 ) , cos2 , or to the given sin cos sin cos sin cos result 2cot2 . Candidates who used Way 1, as described in the mark scheme, gained some credit for rationalising cos3 sin3 to give cos3 cos sin3 sin . Of those candidates sin cos sin cos 1 sin cos sin 2 who correctly applied to the denominator, only a few realised that 2 A common error in using Way 1 was to the numerator could be written as cos2 . 2 2 cos 3 sin 3 simplify cos3 sin3 or cos4 sin 4 . to give sin cos sin cos sin cos Way 2, as described in the mark scheme (and variations of this), was the most popular approach among candidates who completed the proof successfully. Candidates only started to access the marks for Way 2 when they combined their fractions by rationalising the denominator. Errors using Way 2 included the incorrect expansion of cos(2 ) or sin(2 ) , bracketing errors, manipulation errors or sign errors. They usually progressed as far as A few candidates started their proof from 2cot2 . 2 2(1 tan ) 2cos2 2cot 2 , 2cot 2 writing either or but could make no sin 2 2tan creditable progress until they applied a Way 3 method as described in the mark scheme. Some candidates spent a considerable amount of time attempting to write 3 3 4cos 3cos 3sin 4sin cos3 sin3 in terms of to give e.g. . sin cos sin cos candidates only started to access the marks for this method when they combined their fractions by rationalising the denominator. In Q13(b), most candidates used the result given in Q13(a) to deduce the equation 1 2cot 2 4. tan 2 , Most of these then attempted to solve and some obtained the 2 Common errors included attempting to solve tan2 correct answer 103.3 . 2 A few candidates obtained by . and finding more than one value for 103.3 2 2(1 tan ) 4. solving the equation 2tan

Common mistakes

• 2cot2θ = 4 so tan2θ = ½
• 2θ in (180°, 360°) gives only θ = 103.3°
• solving tan2θ = 2 or giving extra solutions

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall this question was well done, with many students taking advantage of being able to attempt (b) without completing (a). The few who restarted unnecessarily in (b) lost valuable time in the examination. In (b) students generally managed to solve the quadratic, often using a calculator to find the two roots. The most common error was to give the signs of the roots the wrong way around, even after correct factorisation in some cases. Almost all found solutions for 3x but many failed to divide by 3. A substantial number of students who went on to find values for x missed a solution (often the 80o) or gave extra solutions outside the range e.g. 103.9°

Common mistakes

• solve 6c² − c − 2 = 0: cos 3x = 2/3 or −1/2 (signs of the roots often reversed)
• find 3x in 0 ≤ 3x < 270° then divide by 3
• x = 16.1°, 40°, 80° only (80° often missed
• 103.9° is outside the range)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question involved modelling the depth of water in a harbour using a trigonometric Most students also started part (b) correctly and were able to proceed to the intermediate t point of sin30 0.6 without much difficulty. The final part of the question, however, proved to be much more difficult, with many students not finding a value of t greater than 8.5, giving 𝑡= 7.2 then a time of 7: 14𝑎𝑚. Of those who correctly found 𝑡= 10.77 a large number failed to give their final answer as a time, with 10 hours and 46 minutes being common. A number of successful students were noteworthy in terms of the clear presentation and communication of their solutions, often producing sketch graphs of the sine curve to aid their understanding.

Common mistakes

• sin(30t)° = −0.6
• need the first t after 8.5 (loading ends 8:30): 30t = 323.1 so t = 10.77, not 7.2 (7:14 am)
• answer as a time 10:46 am, not 10 hours 46 minutes

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was another modelling question, this time modelling the height of a roller coaster above t marks in (c), but many were aware that it would occur when sin(0.25 ) The most 1. or else use 450 rather than 5 thus , , common errors were to find the first value using 2 2 arriving at the incorrect answer of 1800 seconds.

Common mistakes

• second maximum when 0.25t = 5π/2 (radians), T = 10π ≈ 31.4 s
• using π/2 (first maximum) or 450 in degrees (giving 1800) is wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (c), some students correctly deduced that the given equation had 3 real solutions for 7 10 , with a few students incorrectly deducing that there were 6 real solutions. Most students misunderstood the requirement of the question by finding 23.1, 26.2, 29.4, with those students who did not go on to state that there were 3 real solutions losing the mark for this part.

Common mistakes

• equation is −f(tan θ) = 0 so tan θ = 2 only
• one solution per period π gives 3 solutions in 7π ≤ θ < 10π
• listing 23.1, 26.2, 29.4 without stating 3 scores B0
• 6 is a common wrong answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (i) was accessible to most students, whereas some students struggled to gain access to part (ii). 1 x In part (i), many students replaced secx with cos x and arrived at 4sin cos A few 1. x x with sin 4 ,x but most used the identity sin2 students replaced 4sin cos to x 2sin cos x x x 1 x sin 2 . give the correct Many students obtained both correct exact solutions in radians, 2 with only a few omitting the second solution 5 . 12 Some students made no attempt at part (ii) while a significant number of other students did not employ a correct strategy for solving 5sin The majority of students, 5cos 2. however, attempted one of several correct strategies. The most popular strategy was to and 2 2 cos sin 1 square both sides of the given equation and apply the identities 21 sin 2 . to achieve A second strategy was to square both sides of sin2 2sin cos 25 to achieve a quadratic equation 2 2 cos sin 1 and apply the identity 5sin 2 5cos in either cos or sin . In both of these strategies, extra (incorrect) solutions were generated as a result of squaring both sides of their equation. Only a few students rejected these extra A third strategy was to rewrite the 61.4 , 208.6 . solutions to leave the correct solutions 50 sin( 45 ) 2 50 cos( 45 ) 2, given equation as or even which were usually A few students applied other correct strategies which led 61.4 , 208.6 . solved to give to solving quadratic equations in either tan or cot .

Common mistakes

• sec x = 1/cos x
• 4 sin x cos x = 1 gives sin 2x = ½
• x = π/12 and 5π/12 (second solution often omitted)
• answers in radians

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (i) was accessible to most students, whereas some students struggled to gain access to part (ii). 1 x In part (i), many students replaced secx with cos x and arrived at 4sin cos A few 1. x x with sin 4 ,x but most used the identity sin2 students replaced 4sin cos to x 2sin cos x x x 1 x sin 2 . give the correct Many students obtained both correct exact solutions in radians, 2 with only a few omitting the second solution 5 . 12 Some students made no attempt at part (ii) while a significant number of other students did not employ a correct strategy for solving 5sin The majority of students, 5cos 2. however, attempted one of several correct strategies. The most popular strategy was to and 2 2 cos sin 1 square both sides of the given equation and apply the identities 21 sin 2 . to achieve A second strategy was to square both sides of sin2 2sin cos 25 to achieve a quadratic equation 2 2 cos sin 1 and apply the identity 5sin 2 5cos in either cos or sin . In both of these strategies, extra (incorrect) solutions were generated as a result of squaring both sides of their equation. Only a few students rejected these extra A third strategy was to rewrite the 61.4 , 208.6 . solutions to leave the correct solutions 50 sin( 45 ) 2 50 cos( 45 ) 2, given equation as or even which were usually A few students applied other correct strategies which led 61.4 , 208.6 . solved to give to solving quadratic equations in either tan or cot .

Common mistakes

• R sin(θ − α) with R = √50, α = 45°
• squaring both sides creates extra roots that must be rejected
• θ = 61.4°, 208.6°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between students of all abilities with part (a) more accessible to students than part (b). Part (b) proved to be discriminating, with some students not realising that they could use the result from part (a). Some students substituted 1 cos2x with tan sin2 x x and then proceeded to cancel tan sin2 x xfrom both sides, while other students substituted tan sin2 x x with 1 cos2x and then proceeded to cancel 1 cos2x from both sides. These two strategies together with applying the identity 2 2 sec 1 tan x x usually led to a 3-term quadratic in tan .x Most students used a correct method, usually by factorising, to solve their quadratic equation in tan x but only a few stated all three correct solutions. As a result of cancelling either tan sin2 x x or 1 cos2x from both sides of their equation, many students missed out the correct solution 0. x Those students who factorised out these terms were usually more successful in finding 0. x Some students did not give their other two solutions in the required form of the exact 4 x and 1.326 x to 3 decimal places.

Common mistakes

• replace (1 − cos 2x) by tan x sin 2x (or vice versa)
• factorise, do not cancel, so that x = 0 is kept
• sec²x = 1 + tan²x gives tan²x − 3tan x − 4 = 0
• x = −π/4 (exact) and 1.326