AS & A Level · Practice questions · Solving trigonometric equations

Find them all.

Six original Cambridge-style questions on solving trigonometric equations in a given interval, from principal values to quadratics that need an identity.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Principal value, symmetry, period, interval.

One function: use \(\tan x = \dfrac{\sin x}{\cos x}\) or \(\sin^2 x + \cos^2 x = 1\) so only one trig function is left; factorise rather than cancel.
Bracket interval: for \(\sin 2x\) or \(\cos(x + k)\), rewrite the interval for the whole bracket.
Principal value: \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) on the calculator, in the right mode.
Second value: \(180^\circ - \alpha\) (sin), \(360^\circ - \alpha\) (cos), \(\alpha + 180^\circ\) (tan).
Period: add or subtract \(360^\circ\) (tan: \(180^\circ\)) until you leave the interval, then convert back to \(x\).

Recall

[3 marks]

Find the exact values of (a) \(\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)\) (b) \(\cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)\) (c) \(\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\).

(a) \(-\dfrac{\pi}{3}\) (the range of \(\sin^{-1}\) is \(-\dfrac{\pi}{2}\) to \(\dfrac{\pi}{2}\)) ✓
(b) \(\pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}\) (the range of \(\cos^{-1}\) is 0 to \(\pi\)) ✓
(c) \(\dfrac{\pi}{6}\) ✓

Calculation

[3 marks]

Solve the equation \(5\cos x + 2 = 0\) for \(0^\circ \le x \le 360^\circ\).

\(\cos x = -0.4\) ✓
\(x = \cos^{-1}(-0.4) = 113.6^\circ\) ✓
\(x = 360^\circ - 113.6^\circ = 246.4^\circ\) ✓

Calculation

[4 marks]

Solve the equation \(5\sin 2\theta = 2\) for \(0 \le \theta \le \pi\), giving your answers in radians correct to 3 significant figures.

\(\sin 2\theta = 0.4\), with \(0 \le 2\theta \le 2\pi\). ✓
\(2\theta = \sin^{-1}0.4 = 0.41152\) or \(2\theta = \pi - 0.41152 = 2.73008\). ✓
(Adding \(2\pi\) to either exceeds \(2\pi\).) ✓
\(\theta = 0.206\) or \(\theta = 1.37\) ✓

Calculation

[4 marks]

Solve the equation \(2\tan^2 x - \tan x - 6 = 0\) for \(0^\circ \le x \le 180^\circ\).

Factorise: \((2\tan x + 3)(\tan x - 2) = 0\), so \(\tan x = 2\) or \(\tan x = -1.5\). ✓
\(\tan x = 2\): \(x = 63.4^\circ\) (\(243.4^\circ\) is outside the interval). ✓
\(\tan x = -1.5\): principal value \(-56.3^\circ\), add \(180^\circ\): \(x = 123.7^\circ\). ✓
\(x = 63.4^\circ\) or \(123.7^\circ\) ✓

Proof

[5 marks]

(a) Show that the equation \(4\tan\theta\sin\theta = 15\) can be written as \(4\cos^2\theta + 15\cos\theta - 4 = 0\).

(b) Hence solve \(4\tan\theta\sin\theta = 15\) for \(0^\circ \le \theta \le 360^\circ\).

(a) \(4 \times \dfrac{\sin\theta}{\cos\theta} \times \sin\theta = 15\), so \(4\sin^2\theta = 15\cos\theta\). ✓
\(4(1 - \cos^2\theta) = 15\cos\theta\), so \(4\cos^2\theta + 15\cos\theta - 4 = 0\). ✓
(b) \((4\cos\theta - 1)(\cos\theta + 4) = 0\); \(\cos\theta = -4\) is impossible, so \(\cos\theta = \tfrac{1}{4}\). ✓
\(\theta = 75.5^\circ\) or \(360^\circ - 75.5^\circ = 284.5^\circ\) ✓

Analysis

[6 marks]

The temperature in a greenhouse \(t\) hours after midnight is modelled by \(T = 14 - 6\cos\left(\dfrac{\pi t}{12}\right)\) degrees Celsius, for \(0 \le t \le 24\).

(a) Find the exact temperature at 3 a.m.

(b) Find the times at which the temperature is \(17^\circ\text{C}\).

(a) \(t = 3\): \(T = 14 - 6\cos\dfrac{\pi}{4} = 14 - 6 \times \dfrac{\sqrt{2}}{2} = 14 - 3\sqrt{2}\ ^\circ\text{C}\) (about 9.76). ✓
(b) \(14 - 6\cos\dfrac{\pi t}{12} = 17\) gives \(\cos\dfrac{\pi t}{12} = -\tfrac{1}{2}\), with \(0 \le \dfrac{\pi t}{12} \le 2\pi\). ✓
\(\dfrac{\pi t}{12} = \dfrac{2\pi}{3}\) or \(\dfrac{4\pi}{3}\), so \(t = 8\) or \(t = 16\): 8 a.m. and 4 p.m. ✓
(c) \(T\) rises from its minimum \(8^\circ\text{C}\) at midnight to \(20^\circ\text{C}\) at noon, so \(T \ge 17\) for \(8 \le t \le 16\): 8 hours. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.