NECO 2023 · Paper 1 · Q14

Solve $\sin^2x + 2\sin x + 1 = 0$ for $0^\circ < x < 360^\circ$ .

Worked solution (try it first)

The left side is a perfect square: $\sin^2 x + 2\sin x + 1 = (\sin x + 1)^2$ .
So $(\sin x + 1)^2 = 0$ , which gives $\sin x = -1$ .
Between $0^\circ$ and $360^\circ$ this happens only at the bottom of the sine curve, $x = 270^\circ$ , option D.