WAEC 2022 · Paper 2 · Q4

Trigonometry

(a)
Solve $3\cos2x - \sin x = 0$ for $0^\circ \le x \le 360^\circ$ (2 d.p.).
Separate values with commas, e.g. 3, −2

Try it on a graph

x in degrees: the curve y = 3 cos 2x − sin x crosses zero four times.

Worked solution (try it first)

Use $\cos 2x = 1 - 2\sin^2 x$ : $3(1 - 2\sin^2 x) - \sin x = 0$ .
So $6\sin^2 x + \sin x - 3 = 0$ .
$\sin x = \dfrac{-1 \pm \sqrt{1 + 72}}{12}$
$= \dfrac{-1 \pm \sqrt{73}}{12}$ , so $\sin x \approx 0.6287$ or $-0.7953$ .
$\sin x = 0.6287$ : $x \approx 38.95^\circ$ or $180^\circ - 38.95^\circ = 141.05^\circ$ .
$\sin x = -0.7953$ : the reference angle is $52.69^\circ$ , so $x \approx 232.69^\circ$ or $307.31^\circ$ .

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