WAEC 2022 · Paper 1 · Q30

Solve: $4\sin^2\theta + 1 = 2$ , where $0^\circ < \theta < 180^\circ$ .

Worked solution (try it first)

Take 1 from both sides and divide by 4: $\sin^2\theta = \frac14$ .
Square root: $\sin\theta = \frac12$ (sine is positive between $0^\circ$ and $180^\circ$ , so $-\frac12$ is not used).
$\sin\theta = \frac12$ at $\theta = 30^\circ$ and at $180^\circ - 30^\circ = 150^\circ$ , option B.