IGCSE · Practice questions · Solving trigonometric equations

Find them all.

Extended. Six original Cambridge-style questions on solving trigonometric equations for 0° to 360°.

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

Rearrange, inverse, then use the graph.

Rearrange to \(\sin x = k\), \(\cos x = k\) or \(\tan x = k\).
Check that \(-1 \le k \le 1\) for sine and cosine; otherwise no solutions.
First angle from \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\).
Other angles: \(180^\circ - \alpha\) (sin), \(360^\circ - \alpha\) (cos), \(\alpha + 180^\circ\) (tan).
Range: keep only \(0^\circ \le x \le 360^\circ\); give answers to 1 d.p.

Recall

[2 marks]

Without a calculator, solve \(\sin x = \dfrac{1}{2}\) for \(0^\circ \le x \le 360^\circ\).

\(\sin 30^\circ = \dfrac{1}{2}\), so \(x = 30^\circ\). ✓
Second solution: \(180^\circ - 30^\circ = 150^\circ\). ✓

Calculation

[2 marks]

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

\(x = \cos^{-1}(0.35) = 69.5^\circ\) ✓
\(360^\circ - 69.5^\circ = 290.5^\circ\) ✓

Calculation

[3 marks]

Solve \(3\tan x - 4 = 0\) for \(0^\circ \le x \le 360^\circ\).

\(\tan x = \dfrac{4}{3}\) ✓
\(x = \tan^{-1}\left(\dfrac{4}{3}\right) = 53.1^\circ\) ✓
\(53.1^\circ + 180^\circ = 233.1^\circ\) ✓

Calculation

[3 marks]

Solve \(4\sin x + 3 = 0\) for \(0^\circ \le x \le 360^\circ\).

\(\sin x = -0.75\); the calculator gives \(-48.6^\circ\), outside the range. ✓
Acute angle with \(\sin = 0.75\) is \(48.6^\circ\); sine is negative from \(180^\circ\) to \(360^\circ\). ✓
\(x = 180^\circ + 48.6^\circ = 228.6^\circ\) or \(x = 360^\circ - 48.6^\circ = 311.4^\circ\) ✓

Analysis

[3 marks]

(a) Explain why the equation \(2\cos x = 3\) has no solutions.

(b) Without a calculator, solve \(2\cos x = -\sqrt{3}\) for \(0^\circ \le x \le 360^\circ\).

(a) \(\cos x = 1.5\), but \(\cos x\) can never be greater than 1. ✓
(b) \(\cos x = -\dfrac{\sqrt{3}}{2}\). Since \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\) and cosine is negative from \(90^\circ\) to \(270^\circ\): ✓
\(x = 180^\circ - 30^\circ = 150^\circ\) or \(x = 180^\circ + 30^\circ = 210^\circ\) ✓

Analysis

[4 marks]

The depth of water, \(d\) metres, at a harbour entrance is modelled by \(d = 6 + 3\sin x\), where \(x\) is measured in degrees and \(0 \le x \le 360\) covers one tidal cycle.

(a) Find the greatest depth.

(b) A ship needs a depth of at least 7.5 m. Find the values of \(x\) between which the ship can safely enter.

(a) Greatest when \(\sin x = 1\): \(d = 6 + 3 = 9\) m. ✓
(b) Solve \(6 + 3\sin x = 7.5\): \(\sin x = 0.5\). ✓
\(x = 30^\circ\) or \(x = 180^\circ - 30^\circ = 150^\circ\). ✓
Between these the sine graph is above 0.5, so the ship can enter for \(30 \le x \le 150\). ✓

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.