Solve $\cos 2x - 3\sin x - 1 = 0$ using addition formula 
Solve $\cos 2x - 3\sin x - 1 = 0, \quad 0^{\circ} \le x \le 360^{\circ}$

\begin{align} \cos 2x - 3\sin x - 1 = 0
 &\iff 1 - 2\sin^2 x - 3\sin x - 1 = 0 \\
 &\iff- 2\sin^2 x - 3\sin x = 0 \\
 &\iff2\sin^2 x + 3\sin x =0\\
 &\iff\sin x(2\sin x + 3) =0 \\
 &\iff\sin x = 0 \lor  2 \sin x + 3= 0
\end{align}
I could go on but the book gives the answer, $0^{\circ}, 180^{\circ}, 360^{\circ}$ and I am mystified as to where these answers have come from.

$\sin x =0 \Leftrightarrow x =\pi n (0^{\circ}, 180^{\circ}, 360^{\circ})$
$2\sin x=-3 \Rightarrow \sin x = -\frac 32 - $impossible
