A-level · Trigonometry

Solving trigonometric equationsExam-style questions

The full method for solving trig equations in a range — quadratics in sin/cos, using sin²+cos²≡1 to convert a mixed equation, and transformed arguments like sin 2x and cos(x − 30°) where the interval must change.

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13 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for $0 \leqslant x < \pi$ , the equation

$\sqrt{3}\tan\left(3x - \dfrac{\pi}{8}\right) = 1$

giving the answers in radians in the form $k\pi$ , where $k$ is a rational constant to be found.

(3 marks)
(b) Solve, for $0 \leqslant \theta < 360^\circ$ , the equation

$4\sin\theta\tan\theta = 3 - \cos\theta$

giving your answers, in degrees, to one decimal place.

(5 marks)
Level 4 · 8 marksTry it
2
In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

$\tan\theta\,(4\sin\theta + 1) = 6\cos\theta$

can be written as

$10\sin^2\theta + \sin\theta - 6 = 0$

(3 marks)
(b) Hence solve, for $-\pi < x < \pi$ , the equation

$\tan\left(x - \dfrac{\pi}{5}\right)\left(4\sin\left(x - \dfrac{\pi}{5}\right) + 1\right) = 6\cos\left(x - \dfrac{\pi}{5}\right)$

giving your answers to 3 significant figures.

(4 marks)
Level 4 · 7 marksTry it
3
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for $-\pi < x < \pi$ , the equation

$6\cos(2x - 0.3) + 1 = 0$

giving your answers in radians to 2 decimal places.

(4 marks)
(b) Solve, for $-180^\circ < \theta < 180^\circ$ , the equation

$3\sin\theta\tan\theta - 5\cos\theta = 2$

giving your answers, in degrees, to one decimal place.

(5 marks)
Level 4 · 9 marksTry it
4
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

$4\cos\theta\,(2 - \tan\theta\sin\theta) = 2 + 3\cos\theta$

may be written as

$4\cos^2\theta + 5\cos\theta - 6 = 0$

(3 marks)
(b) Hence solve, for $0 < x < 180^\circ$

$4\cos 3x\,(2 - \tan 3x\sin 3x) = 2 + 3\cos 3x$

giving your answers to one decimal place.

(4 marks)
Level 4 · 7 marksTry it
5
Solve the equation

$2\cos\left(2x - \dfrac{\pi}{3}\right) = 1$

for $0 \leqslant x \leqslant \pi$ , giving your answers in terms of $\pi$ . (5 marks)
Level 4 · 5 marksTry it