Solving trigonometric equations - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Solving trigonometric equations for Edexcel A-Level Maths (Paper 2). 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.

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2–9 marks each
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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

3 marks

Solve, for \(0 \leqslant x \leqslant 180^\circ\), the equation
\[10\sin x\cos x = 3\]
Give your answers to one decimal place.

Model answer

\(10\sin x\cos x = 5\sin 2x\), so \(5\sin 2x = 3\), \(\sin 2x = 0.6\)
For \(0 \leqslant x \leqslant 180^\circ\): \(0 \leqslant 2x \leqslant 360^\circ\)
\(2x = 36.87^\circ\) or \(180^\circ - 36.87^\circ = 143.13^\circ\)
\(x = 18.4^\circ\) or \(x = 71.6^\circ\)

Explanation

The product \(\sin x\cos x\) is half of \(\sin 2x\), so the equation becomes \(\sin 2x = 0.6\). Doubling the interval for 2x gives two values, which are then halved. A common error, noted by examiners, is to replace a multiple of \(\sin x\cos x\) by a sine of a larger multiple of x, such as \(\sin 10x\).

Mark scheme

1 mark
M1 Uses \(2\sin x\cos x = \sin 2x\) to obtain \(5\sin 2x = 3\), i.e. \(\sin 2x = 0.6\); or correct final answers of \(x = 18.4^\circ\) and \(x = 71.6^\circ\) with no working
An answer of x = 18.4° and x = 71.6° with no working scores 3 marks (awrt, i.e. in [18.35, 18.45) and [71.55, 71.65)). \(10\sin x\cos x = \sin 10x\) or \(= 10\sin 2x\) scores M0.
1 mark
M1 (dM1) Finds two values of 2x in the range \(0 \leqslant 2x \leqslant 360^\circ\): \(36.87^\circ\) and \(180^\circ - 36.87^\circ = 143.13^\circ\), and halves
Depends on the previous M mark. \(\arcsin 0.6 = 36.87^\circ\) (accept awrt 36.9, i.e. in [36.85, 36.95)).
1 mark
A1 \(x = 18.4^\circ\) and \(x = 71.6^\circ\) only
Accept awrt 18.4 and 71.6, i.e. in [18.35, 18.45) and [71.55, 71.65). Withhold for extra values in the range.

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 \(-\pi \leqslant \theta \leqslant \pi\), the equation
\[8\sin^4\theta + 1 = 6\sin^2\theta\]
Give your answers as exact multiples of \(\pi\).

Model answer

\(8\sin^4\theta - 6\sin^2\theta + 1 = 0\)
\((4\sin^2\theta - 1)(2\sin^2\theta - 1) = 0\)
\(\sin^2\theta = \dfrac{1}{4}\Rightarrow \sin\theta = \pm\dfrac{1}{2} \Rightarrow \theta = \pm\dfrac{\pi}{6}, \pm\dfrac{5\pi}{6}\)
\(\sin^2\theta = \dfrac{1}{2}\Rightarrow \sin\theta = \pm\dfrac{1}{\sqrt{2}} \Rightarrow \theta = \pm\dfrac{\pi}{4}, \pm\dfrac{3\pi}{4}\)
\(\theta = -\dfrac{5\pi}{6}, -\dfrac{3\pi}{4}, -\dfrac{\pi}{4}, -\dfrac{\pi}{6}, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{6}\)

Explanation

The equation is a quadratic in \(\sin^2\theta\), which factorises to give \(\sin^2\theta = \dfrac{1}{4}\) or \(\dfrac{1}{2}\). Each value gives a positive and a negative square root, and each of these gives two angles in \(-\pi \leqslant \theta \leqslant \pi\), so there are eight solutions. The common error is forgetting the negative square roots, which loses the four negative solutions.

Mark scheme

1 mark
M1 Treats the equation as a quadratic in \(\sin^2\theta\), \(8\sin^4\theta - 6\sin^2\theta + 1 = 0\), and solves it by factorising, the formula or completing the square. An answer with no working scores 0 marks.
An answer of \(\pm\dfrac{\pi}{6}, \pm\dfrac{\pi}{4}, \pm\dfrac{3\pi}{4}, \pm\dfrac{5\pi}{6}\) with no working scores 0 marks: the question demands working. Alternative method: \(\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}\) gives \(2\cos^2 2\theta - \cos 2\theta = 0\), so \(\cos 2\theta = 0\) or \(\dfrac{1}{2}\): M1 A1 for this, then the same marks.
1 mark
A1 \((4\sin^2\theta - 1)(2\sin^2\theta - 1) = 0\), so \(\sin^2\theta = \dfrac{1}{4}\) or \(\sin^2\theta = \dfrac{1}{2}\)
Both values needed.
1 mark
M1 (dM1) Takes square roots, including the negative roots, \(\sin\theta = \pm\dfrac{1}{2}\) or \(\sin\theta = \pm\dfrac{1}{\sqrt{2}}\), and finds at least one correct value of \(\theta\)
Depends on the first M mark (point 1). Taking only the positive roots is M1 here but loses the A marks for the negative solutions.
1 mark
A1 \(\theta = \pm\dfrac{\pi}{6}, \pm\dfrac{5\pi}{6}\)
All four, exact. Degrees (±30°, ±150°) score A0.
1 mark
A1 \(\theta = \pm\dfrac{\pi}{4}, \pm\dfrac{3\pi}{4}\), and no other values
Depends on the M mark in point 3. All four, exact; withhold for any extra value in the interval.

Question 3

6 marks

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

Show that
\[\cos^2\left(x + \dfrac{\pi}{4}\right) - \sin^2\left(x + \dfrac{\pi}{4}\right) \equiv -\sin 2x\]
Hence solve, for \(0 \leqslant x < 2\pi\), the equation
\[\cos^2\left(x + \dfrac{\pi}{4}\right) - \sin^2\left(x + \dfrac{\pi}{4}\right) = 0.4\]
giving your answers in radians to 3 significant figures.

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

Model answer

\(\cos^2 A - \sin^2 A = \cos 2A\), with \(A = x + \dfrac{\pi}{4}\):
LHS \(= \cos\left(2x + \dfrac{\pi}{2}\right) = \cos 2x\cos\dfrac{\pi}{2} - \sin 2x\sin\dfrac{\pi}{2} = 0 - \sin 2x = -\sin 2x\), as required.
Hence \(-\sin 2x = 0.4\), so \(\sin 2x = -0.4\)
For \(0 \leqslant x < 2\pi\): \(0 \leqslant 2x < 4\pi\)
\(\arcsin 0.4 = 0.4115\)
\(2x = \pi + 0.4115, 2\pi - 0.4115, 3\pi + 0.4115, 4\pi - 0.4115 = 3.5531, 5.8717, 9.8363, 12.1549\)
\(x = 1.78, 2.94, 4.92, 6.08\) (3 s.f.)

Explanation

The left-hand side has the form \(\cos^2 A - \sin^2 A\), which is \(\cos 2A\), and \(\cos\left(2x + \dfrac{\pi}{2}\right) = -\sin 2x\). The equation then becomes \(\sin 2x = -0.4\), and since 2x runs from 0 to \(4\pi\) there are four solutions, all in the third and fourth quadrants of each turn. Common errors are a sign slip giving \(\sin 2x = 0.4\) and finding only two solutions.

Mark scheme

1 mark
M1 Uses \(\cos^2 A - \sin^2 A = \cos 2A\) with \(A = x + \dfrac{\pi}{4}\) to obtain \(\cos\left(2x + \dfrac{\pi}{2}\right)\). An answer with no working scores 0 marks.
An answer of x = 1.78, 2.94, 4.92, 6.08 with no working scores 0 marks: the question demands working. Alternative: expand \(\cos\left(x + \dfrac{\pi}{4}\right)\) and \(\sin\left(x + \dfrac{\pi}{4}\right)\) using the compound angle formulae and square; M1 for a correct expansion of both.
1 mark
A1* \(\cos\left(2x + \dfrac{\pi}{2}\right) = \cos 2x\cos\dfrac{\pi}{2} - \sin 2x\sin\dfrac{\pi}{2} = -\sin 2x\) cso
The step showing \(\cos\dfrac{\pi}{2} = 0\) and \(\sin\dfrac{\pi}{2} = 1\) (or an equivalent reason) must be seen. In the alternative: \(\dfrac{1}{2}(\cos x - \sin x)^2 - \dfrac{1}{2}(\sin x + \cos x)^2 = -2\sin x\cos x = -\sin 2x\).
1 mark
M1 Uses the identity: \(-\sin 2x = 0.4\), so \(\sin 2x = -0.4\), and finds one value of 2x: \(\arcsin(-0.4) = -0.4115\) leading to a value in \(0 \leqslant 2x < 4\pi\), e.g. \(\pi + 0.4115\)
"Hence" requires the identity to be used. Using \(\sin 2x = 0.4\) scores M0. \(\arcsin 0.4 = 0.4115\) (accept awrt 0.412, i.e. in [0.4115, 0.4125)).
1 mark
M1 (dM1) Finds four values of 2x in the range \(0 \leqslant 2x < 4\pi\) (\(\pi + 0.4115\), \(2\pi - 0.4115\), \(3\pi + 0.4115\), \(4\pi - 0.4115\)) and halves
Depends on the previous M mark.
1 mark
A1 Two correct values from \(x =\) awrt 1.78, 2.94, 4.92, 6.08
i.e. in [1.775, 1.785), [2.935, 2.945), [4.915, 4.925), [6.075, 6.085).
1 mark
A1 All four values \(x = 1.78, 2.94, 4.92, 6.08\) and no others in the range
Depends on the M mark in point 4. Exact values 1.7766, 2.9358, 4.9181, 6.0774. Degrees score A0.

Question 4

7 marks

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

A guitar string \(AB\) of length 65 cm is fixed at both ends. At one instant during its vibration, the displacement, \(y\) mm, of the point of the string at a distance \(x\) cm from \(A\) is modelled by
\[y = 0.3\sin\left(\dfrac{\pi x}{65}\right) + 0.2\sin\left(\dfrac{2\pi x}{65}\right), \quad 0 \leqslant x \leqslant 65\]
where the angles are in radians.

Using a double angle formula, find the distance from \(A\) of the point of the string, other than \(A\) and \(B\), at which the displacement is zero, giving your answer to 3 significant figures.

Find also the distance from \(A\) of the point, other than \(A\) and \(B\), at which the two terms in the model have equal values, giving your answer to 3 significant figures.

Model answer

Let \(u = \dfrac{\pi x}{65}\), so \(0 \leqslant u \leqslant \pi\).
\(y = 0.3\sin u + 0.2\sin 2u = 0.3\sin u + 0.4\sin u\cos u = \sin u(0.3 + 0.4\cos u)\)
\(y = 0\): \(\sin u = 0\) gives u = 0 or π, i.e. x = 0 or 65, the ends A and B.
Otherwise \(\cos u = -\dfrac{0.3}{0.4} = -\dfrac{3}{4}\), \(u = 2.4189\)
\(x = \dfrac{65 \times 2.4189}{\pi} = 50.0\) cm
Equal terms: \(0.3\sin u = 0.4\sin u\cos u\); for \(0 < u < \pi\), \(\sin u ≠ 0\), so \(\cos u = \dfrac{3}{4}\)
\(u = 0.7227\), \(x = \dfrac{65 \times 0.7227}{\pi} = 15.0\) cm

Explanation

With \(u = \dfrac{\pi x}{65}\), the double angle formula shows that both terms share the factor \(\sin u\), which is zero only at the fixed ends. The remaining factors give \(\cos u = -\dfrac{3}{4}\) for zero displacement and \(\cos u = \dfrac{3}{4}\) for equal terms, and each has one solution with u between 0 and π. Common errors are cancelling \(\sin u\) without comment and forgetting to convert u back to x.

Mark scheme

1 mark
M1 Uses \(\sin 2u = 2\sin u\cos u\), where \(u = \dfrac{\pi x}{65}\), to write \(y = 0.3\sin u + 0.4\sin u\cos u\). An answer with no working scores 0 marks.
An answer of 50.0 cm and 15.0 cm with no working scores 0 marks: the question demands working. (Accepted ranges, when working is seen: [49.95, 50.05) and [14.95, 15.05).)
1 mark
M1 (dM1) Sets \(y = 0\) and factorises: \(\sin u(0.3 + 0.4\cos u) = 0\)
Depends on the previous M mark. Cancelling \(\sin u\) is acceptable here, provided \(\sin u = 0\) is dealt with (point 3).
1 mark
B1 \(\sin u = 0\) gives \(x = 0\) or \(x = 65\), the fixed ends A and B
Or a statement that \(\sin u ≠ 0\) for \(0 < x < 65\) because \(0 < u < \pi\).
1 mark
A1 \(\cos u = -\dfrac{3}{4}\), so \(u = 2.4189\) and \(x = \dfrac{65u}{\pi} =\) awrt 50.0 (cm), i.e. in [49.95, 50.05)
Depends on the M mark in point 2. Exact value 50.047. Do not accept 50 alone unless 50.0 or more accurate working is seen. Working in degrees: \(u = 138.59^\circ\) and \(x = \dfrac{65 \times 138.59}{180}\) is acceptable.
1 mark
M1 Sets \(0.3\sin u = 0.2\sin 2u = 0.4\sin u\cos u\) and divides by \(\sin u\) (non-zero between A and B) or factorises
The coefficients 0.3, 0.2 and 0.4 are exact. Dividing by \(\sin u\) is valid here because \(\sin u ≠ 0\) for \(0 < u < \pi\); factorising is also accepted.
1 mark
A1 \(\cos u = \dfrac{3}{4}\)
Accept 0.75.
1 mark
A1 \(x =\) awrt 15.0 (cm), i.e. in [14.95, 15.05)
Depends on the M mark in point 5. From \(u = \arccos 0.75 = 0.7227\), \(x = \dfrac{65 \times 0.7227}{\pi} = 14.953\). Do not accept 15 alone unless 14.95 or more accurate working is seen.

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