Solving trigonometric equations - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Solving trigonometric equations for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
3–9 marks each
Instant marking
No sign-up needed

18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

4 marks

Solve, for \(0 \leqslant x < \pi\), the equation
\[\cos\left(3x - \dfrac{\pi}{4}\right) = -0.4\]
Give your answers in radians to 3 significant figures.

Model answer

For \(0 \leqslant x < \pi\): \(-\dfrac{\pi}{4} \leqslant 3x - \dfrac{\pi}{4} < \dfrac{11\pi}{4}\), i.e. \(-0.785 \leqslant 3x - \dfrac{\pi}{4} < 8.639\)
\(\arccos(-0.4) = 1.9823...\)
\(3x - \dfrac{\pi}{4} = 1.9823..., \ 2\pi - 1.9823... = 4.3009..., \ 2\pi + 1.9823... = 8.2655...\)
\(3x = 2.7677..., \ 5.0863..., \ 9.0509...\)
\(x = 0.923, 1.70, 3.02\) (3 s.f.)

Explanation

Treat \(3x - \dfrac{\pi}{4}\) as a single angle, which runs from \(-\dfrac{\pi}{4}\) to \(\dfrac{11\pi}{4}\) when x runs from 0 to \(\pi\). Cosine is negative in the second and third quadrants, so the values are \(1.982\), \(2\pi - 1.982\) and \(2\pi + 1.982\); adding \(\dfrac{\pi}{4}\) and dividing by 3 gives x. Common errors are dividing by 3 before adding \(\dfrac{\pi}{4}\) and not extending the interval for the new angle.

Mark scheme

1 mark
B1 Principal value \(3x - \dfrac{\pi}{4} = \arccos(-0.4) = 1.982...\); or a correct final answer of \(x = 0.923, 1.70, 3.02\) with no working (scores all 4 marks)
An answer of x = 0.923, 1.70, 3.02 with no working scores 4 marks (the question does not demand working). With no working, an incomplete answer that includes 0.923 scores B1 M0 M1 A0. Accept awrt 1.98, i.e. in [1.975, 1.985), for the principal value.
1 mark
M1 Finds at least one further value of \(3x - \dfrac{\pi}{4}\) in the interval \(-\dfrac{\pi}{4} \leqslant 3x - \dfrac{\pi}{4} < \dfrac{11\pi}{4}\): \(2\pi - 1.982... = 4.300...\) or \(2\pi + 1.982... = 8.265...\)
Accept awrt 4.30 and awrt 8.27, i.e. in [4.295, 4.305) and [8.265, 8.275). The value \(-1.982\) is outside the interval and must not be used.
1 mark
M1 Adds \(\dfrac{\pi}{4}\) and then divides by 3 for at least one value
Dividing by 3 before adding \(\dfrac{\pi}{4}\) is M0.
1 mark
A1 \(x =\) awrt 0.923, 1.70, 3.02 (each in [0.9225, 0.9235), [1.695, 1.705), [3.015, 3.025)) and no others in the range
All three needed; extra values in the range lose this mark. Answers in degrees (52.9°, 97.1°, 172.9°) score A0. A common error is to use only \(0 \leqslant 3x - \dfrac{\pi}{4} < 2\pi\) and so miss 3.02.

Question 2

5 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Solve, for \(0 \leqslant x < 360^\circ\), the equation
\[\sin(x + 60^\circ) = 2\sin x\]
Give your answers in degrees.

[You may use \(\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B\)]

Model answer

\(\sin x\cos 60^\circ + \cos x\sin 60^\circ = 2\sin x\)
\(\dfrac{1}{2}\sin x + \dfrac{\sqrt{3}}{2}\cos x = 2\sin x\)
\(\dfrac{\sqrt{3}}{2}\cos x = \dfrac{3}{2}\sin x\)
\(\tan x = \dfrac{\sqrt{3}}{3} = \dfrac{1}{\sqrt{3}}\)
\(x = 30^\circ\) or \(x = 30^\circ + 180^\circ = 210^\circ\)

Explanation

Expanding \(\sin(x + 60^\circ)\) with the exact values \(\cos 60^\circ = \dfrac{1}{2}\) and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\) gives terms in \(\sin x\) and \(\cos x\) only. Collecting them and dividing by \(\cos x\) gives \(\tan x = \dfrac{1}{\sqrt{3}}\), with solutions 180° apart. Common errors are mixing up the exact values of sin 60° and cos 60° and giving only one solution.

Mark scheme

1 mark
M1 Expands \(\sin(x + 60^\circ) = \sin x\cos 60^\circ + \cos x\sin 60^\circ\). An answer with no working scores 0 marks.
An answer of x = 30°, 210° with no working scores 0 marks: the question demands working.
1 mark
A1 \(\dfrac{1}{2}\sin x + \dfrac{\sqrt{3}}{2}\cos x = 2\sin x\) (exact values used)
1 mark
M1 (dM1) Collects the sine terms and divides by \(\cos x\) to reach \(\tan x = k\): \(\dfrac{\sqrt{3}}{2}\cos x = \dfrac{3}{2}\sin x\)
Depends on the first M mark (not the point just before). Dividing by \(\cos x\) is valid because \(\cos x = 0\) does not satisfy the equation.
1 mark
A1 \(\tan x = \dfrac{1}{\sqrt{3}}\) (or \(\dfrac{\sqrt{3}}{3}\))
Accept awrt 0.577, i.e. in [0.5765, 0.5775), if the next mark's angles are exact.
1 mark
A1 \(x = 30^\circ, 210^\circ\) and no others in the range
Accept 30.0° and 210.0°. Answers in radians score A0. A common error is a slip with the exact values, e.g. \(\cos 60^\circ = \dfrac{\sqrt{3}}{2}\).

Question 3

6 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Solve, for \(0 \leqslant \theta < 360^\circ\), the equation
\[\tan 2\theta = 4\tan\theta\]
Give your answers to one decimal place where they are not exact.

Model answer

\(\dfrac{2\tan\theta}{1 - \tan^2\theta} = 4\tan\theta\)
\(2\tan\theta = 4\tan\theta - 4\tan^3\theta\)
\(4\tan^3\theta - 2\tan\theta = 0\)
\(2\tan\theta(2\tan^2\theta - 1) = 0\)
\(\tan\theta = 0 \Rightarrow \theta = 0^\circ, 180^\circ\)
\(\tan\theta = \dfrac{1}{\sqrt{2}} \Rightarrow \theta = 35.3^\circ, 215.3^\circ\)
\(\tan\theta = -\dfrac{1}{\sqrt{2}} \Rightarrow \theta = 144.7^\circ, 324.7^\circ\)

Explanation

Replace \(\tan 2\theta\) with \(\dfrac{2\tan\theta}{1 - \tan^2\theta}\) and multiply up to get a cubic in \(\tan\theta\) with a common factor. Factorising keeps the solutions from \(\tan\theta = 0\), and the quadratic factor gives \(\tan\theta = \pm\dfrac{1}{\sqrt{2}}\), each with two angles 180° apart. The usual errors are cancelling \(\tan\theta\), which loses 0° and 180°, and forgetting the negative square root.

Mark scheme

1 mark
M1 Uses \(\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}\). An answer with no working scores 0 marks.
An answer of \(\theta = 0^\circ, 35.3^\circ, 144.7^\circ, 180^\circ, 215.3^\circ, 324.7^\circ\) with no working scores 0 marks: the question demands working.
1 mark
A1 \(2\tan\theta = 4\tan\theta(1 - \tan^2\theta)\), or \(4\tan^3\theta - 2\tan\theta = 0\), or equivalent
1 mark
M1 (dM1) Factorises out \(\tan\theta\) (rather than cancelling it), \(2\tan\theta(2\tan^2\theta - 1) = 0\), and solves
Depends on the first M mark (not the point just before). Cancelling \(\tan\theta\) can still score this mark for solving \(2\tan^2\theta - 1 = 0\), but then \(\theta = 0^\circ, 180^\circ\) are lost.
1 mark
A1 \(\tan\theta = 0\) and \(\tan\theta = \pm\dfrac{1}{\sqrt{2}}\)
Accept \(\pm\dfrac{\sqrt{2}}{2}\) or awrt \(\pm 0.707\), i.e. in [0.7065, 0.7075). Both signs needed.
1 mark
A1 \(\theta = 0^\circ\) and \(180^\circ\)
Depends on the second M mark (not the point just before). A common error is cancelling \(\tan\theta\), which loses these two solutions.
1 mark
A1 \(\theta =\) awrt 35.3°, 144.7°, 215.3°, 324.7° (each in [35.25, 35.35), [144.65, 144.75), [215.25, 215.35), [324.65, 324.75)) and no others in the range
Depends on the second M mark (not the point just before). All four needed; extra values in the range lose this mark. Answers in radians score A0.

Question 4

8 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Show that every solution of the equation
\[7\sin x = 5 + \cos x\]
is also a solution of the equation
\[25\cos^2 x + 5\cos x - 12 = 0\]
Hence solve, for \(0 \leqslant x < 360^\circ\), the equation \(7\sin x = 5 + \cos x\), giving your answers to one decimal place. You must explain clearly why any other angles that you find are rejected.

Model answer

\(7\sin x = 5 + \cos x\)
Squaring: \(49\sin^2 x = 25 + 10\cos x + \cos^2 x\)
\(49(1 - \cos^2 x) = 25 + 10\cos x + \cos^2 x\)
\(50\cos^2 x + 10\cos x - 24 = 0\)
\(25\cos^2 x + 5\cos x - 12 = 0\), as required.
\((5\cos x - 3)(5\cos x + 4) = 0\), so \(\cos x = \dfrac{3}{5}\) or \(\cos x = -\dfrac{4}{5}\)
\(\cos x = \dfrac{3}{5}\): \(x = 53.1^\circ\) or \(306.9^\circ\)
\(\cos x = -\dfrac{4}{5}\): \(x = 143.1^\circ\) or \(216.9^\circ\)
Since \(5 + \cos x > 0\), every solution has \(\sin x > 0\).
Check \(x = 53.1^\circ\): \(7 \times 0.8 = 5.6 = 5 + 0.6\), valid; \(x = 143.1^\circ\): \(7 \times 0.6 = 4.2 = 5 - 0.8\), valid.
At \(x = 306.9^\circ\) and \(216.9^\circ\), \(\sin x < 0\), so \(7\sin x\) is negative while \(5 + \cos x\) is positive: these angles were introduced by squaring and are rejected.
\(x = 53.1^\circ, 143.1^\circ\)

Explanation

Squaring both sides removes the mixture of \(\sin x\) and \(\cos x\), and \(\sin^2 x = 1 - \cos^2 x\) then gives a quadratic in \(\cos x\) that factorises. Squaring also creates the solutions of \(7\sin x = -(5 + \cos x)\), so each of the four angles must be checked in the original equation: only \(53.1^\circ\) and \(143.1^\circ\), where \(\sin x > 0\), are valid. Common errors are squaring \(5 + \cos x\) as \(25 + \cos^2 x\) and keeping all four angles.

Mark scheme

1 mark
M1 Squares both sides: \(49\sin^2 x = 25 + 10\cos x + \cos^2 x\). An answer with no working scores 0 marks.
An answer of \(x = 53.1^\circ, 143.1^\circ\) with no working scores 0 marks: the question demands working. Squaring must give three terms on the right; \(49\sin^2 x = 25 + \cos^2 x\) is M0.
1 mark
M1 Uses \(\sin^2 x = 1 - \cos^2 x\)
Not dependent on the first M mark.
1 mark
A1* \(49 - 49\cos^2 x = 25 + 10\cos x + \cos^2 x \Rightarrow 25\cos^2 x + 5\cos x - 12 = 0\), cso
Requires both M marks. Given answer: a line such as \(50\cos^2 x + 10\cos x - 24 = 0\) before the result, and the "= 0", must be seen.
1 mark
M1 Solves the quadratic by factorising, the formula or completing the square: \((5\cos x - 3)(5\cos x + 4) = 0\)
Not dependent on the show-that marks. Roots written straight from a calculator score M0 here because of the instruction in the question.
1 mark
A1 \(\cos x = \dfrac{3}{5}\) and \(\cos x = -\dfrac{4}{5}\)
Both values needed.
1 mark
M1 Finds the angles in the range from at least one of their values of \(\cos x\), e.g. \(53.1^\circ\) and \(306.9^\circ\), or \(143.1^\circ\) and \(216.9^\circ\)
Depends on the M mark for solving the quadratic (not the point just before). An R-form method instead, \(\sqrt{50}\sin(x - 8.13^\circ) = 5\), giving \(53.1^\circ\) and \(143.1^\circ\) and no others, is not the 'hence' method but scores this M1, the next A1 and the B1.
1 mark
A1 \(x = 53.1^\circ\) and \(x = 143.1^\circ\) only (awrt, i.e. in [53.05, 53.15) and [143.05, 143.15))
Extra values in the final answer lose this mark. Accept 53.13 and 143.13.
1 mark
B1 Rejects \(306.9^\circ\) and \(216.9^\circ\) with a reason: they do not satisfy \(7\sin x = 5 + \cos x\), because at these angles \(\sin x < 0\), so \(7\sin x = -5.6\) or \(-4.2\) while \(5 + \cos x = 5.6\) or 4.2; they were introduced by squaring (they solve \(7\sin x = -(5 + \cos x)\))
A check by substitution, or the observation that \(5 + \cos x > 0\) so \(\sin x\) must be positive, is enough. Accept the rejected angles awrt 306.9 and 216.9, i.e. in [306.85, 306.95) and [216.85, 216.95).

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

Study Helper Hints first, answers after you try

Stuck on this one? Ask for a hint and I’ll point you in the right direction, without giving the answer away. Once you’ve checked your answer, I’ll show you how the marks are awarded.