JAMB 1989 · UME · Q42

If $4\sin^2x - 3 = 0$ , find $x$ for $0^\circ < x < 90^\circ$ .

Worked solution (try it first)

Add 3 and divide by 4: $\sin^2 x = \frac34$ .
Take the square root (positive, since $x$ is acute): $\sin x = \frac{\sqrt3}{2}$ .
So $x = 60^\circ$ , option C.