IGCSE · Practice questions · Solving trigonometric equations

Find them all.

Six original Cambridge-style questions on solving trigonometric equations in a given domain: reciprocal functions, multiple and compound angles in degrees and radians, quadratics in one function using the Pythagorean identities, and an identity proof that leads to exact solutions.

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

The method

One function, widen the domain, then use symmetry.

One function: use \(\frac{\sin x}{\cos x} = \tan x\), \(\cos^{2} x = 1 - \sin^{2} x\), \(\sec^{2} x = 1 + \tan^{2} x\) or \(\operatorname{cosec}^{2} x = 1 + \cot^{2} x\).
Factorise, never divide by \(\sin x\), \(\cos x\) or \(\tan x\); reject impossible values such as \(\sin x = 2\) or \(\operatorname{cosec} x = \frac{1}{2}\).
Reciprocals: turn \(\sec\), \(\operatorname{cosec}\), \(\cot\) into \(\cos\), \(\sin\), \(\tan\) of the reciprocal value.
Widen the domain for \(2x\), \(\frac{x}{3}\) or \(2x - c\) before solving, then transform back at the end.
Symmetry: sine \(180^\circ - \alpha\), cosine \(360^\circ - \alpha\), tangent \(180^\circ + \alpha\); then add or subtract whole periods. Degrees to 1 d.p., radians to 3 s.f. or exact.

Recall

[2 marks]

Solve \(\cos x = -0.35\) for \(0^\circ \le x \le 360^\circ\).

Principal value: \(\cos^{-1}(-0.35) = 110.49^\circ\). ✓
Other solution: \(360^\circ - 110.49^\circ = 249.51^\circ\). ✓
\(x = 110.5^\circ\) or \(x = 249.5^\circ\). ✓

Calculation

[3 marks]

Solve \(\tan(2x + 30^\circ) = -1\) for \(0^\circ \le x \le 180^\circ\).

Let \(u = 2x + 30^\circ\). Then \(30^\circ \le u \le 390^\circ\). ✓
\(\tan u = -1\): the principal value \(-45^\circ\) is outside, so add \(180^\circ\): \(u = 135^\circ\) or \(315^\circ\) (the next, \(495^\circ\), is too big). ✓
\(2x = 105^\circ\) or \(285^\circ\), so \(x = 52.5^\circ\) or \(x = 142.5^\circ\). ✓

Calculation

[4 marks]

Solve \(5\cos(2x - 1) = 2\) for \(0 \le x \le \pi\), where \(x\) is in radians. Give your answers to 3 significant figures.

Let \(u = 2x - 1\). Then \(-1 \le u \le 2\pi - 1\), that is \(-1 \le u \le 5.283\). ✓
\(\cos u = 0.4\): \(u = \cos^{-1} 0.4 = 1.1593\) or \(u = 2\pi - 1.1593 = 5.1239\). The value \(-1.1593\) is less than \(-1\), so it is outside. ✓
\(x = \dfrac{u + 1}{2}\): \(x = \dfrac{2.1593}{2} = 1.0796\) or \(x = \dfrac{6.1239}{2} = 3.0620\). ✓
\(x = 1.08\) or \(x = 3.06\). ✓

Analysis

[4 marks]

Solve \(2\cot^{2} x + 3\operatorname{cosec} x = 0\) for \(0^\circ < x < 360^\circ\).

Use \(\cot^{2} x = \operatorname{cosec}^{2} x - 1\): \(2\operatorname{cosec}^{2} x - 2 + 3\operatorname{cosec} x = 0\). ✓
\(2\operatorname{cosec}^{2} x + 3\operatorname{cosec} x - 2 = 0\), so \((2\operatorname{cosec} x - 1)(\operatorname{cosec} x + 2) = 0\). ✓
\(\operatorname{cosec} x = \frac{1}{2}\) would need \(\sin x = 2\): impossible. So \(\operatorname{cosec} x = -2\), \(\sin x = -\frac{1}{2}\). ✓
\(x = 180^\circ + 30^\circ = 210^\circ\) or \(x = 360^\circ - 30^\circ = 330^\circ\). ✓

Analysis

[5 marks]

Solve \(3\sin x\tan x = 8\) for \(0^\circ \le x \le 360^\circ\).

\(\tan x = \dfrac{\sin x}{\cos x}\), so \(\dfrac{3\sin^{2} x}{\cos x} = 8\) and \(3\sin^{2} x = 8\cos x\). ✓
\(3(1 - \cos^{2} x) = 8\cos x\), so \(3\cos^{2} x + 8\cos x - 3 = 0\). ✓
\((3\cos x - 1)(\cos x + 3) = 0\). \(\cos x = -3\) is impossible, so \(\cos x = \frac{1}{3}\). ✓
\(\cos^{-1}\frac{1}{3} = 70.53^\circ\), and \(360^\circ - 70.53^\circ = 289.47^\circ\). ✓
\(x = 70.5^\circ\) or \(x = 289.5^\circ\). ✓

Proof

[6 marks]

(a) Show that \(\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = 2\sec^{2} x\).

(b) Hence solve \(\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = 8\) for \(0 \le x \le 2\pi\), giving your answers as exact multiples of \(\pi\).

(a) Common denominator: \[\frac{(1 + \sin x) + (1 - \sin x)}{(1 - \sin x)(1 + \sin x)} = \frac{2}{1 - \sin^{2} x}\] ✓
\(1 - \sin^{2} x = \cos^{2} x\), so the left side is \(\dfrac{2}{\cos^{2} x} = 2\sec^{2} x\). ✓
(b) \(2\sec^{2} x = 8\), so \(\sec^{2} x = 4\) and \(\cos^{2} x = \frac{1}{4}\), giving \(\cos x = \pm\frac{1}{2}\). ✓
\(\cos x = \frac{1}{2}\): \(x = \frac{\pi}{3}\) or \(\frac{5\pi}{3}\). \(\cos x = -\frac{1}{2}\): \(x = \frac{2\pi}{3}\) or \(\frac{4\pi}{3}\). ✓
\(x = \dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}\) or \(\dfrac{5\pi}{3}\) (none makes \(\sin x = \pm 1\), so every one is valid). ✓

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.