WAEC 2022 · Paper 2 · Q4

(a)
Solve, correct to the nearest degree, $3\cos^2\theta + 10\cos\theta - 8 = 0$ , for $0^\circ \le \theta \le 360^\circ$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Treat it as a quadratic in $\cos\theta$ : $(3\cos\theta - 2)(\cos\theta + 4) = 0$ .
$\cos\theta = -4$ is impossible, so $\cos\theta = \frac23$ .
$\theta = \cos^{-1}\frac23$
$\approx 48.19^\circ$ or $360^\circ - 48.19^\circ = 311.81^\circ$ .
To the nearest degree: $\theta = 48^\circ$ or $312^\circ$ .