## Precalculus (6th Edition) Blitzer

The solutions are $\frac{\pi }{2},\frac{\pi }{6},\frac{5}{6}$.
Let us consider the provided equation, $\cos 2x+3\sin x-2=0$ Since, $\cos 2x=1-2{{\sin }^{2}}x$ So, \begin{align} & \cos 2x+3\sin x-2=0 \\ & 1-2{{\sin }^{2}}x+3\sin x-2=0 \end{align} Multiply both sides by $-1$, to get \begin{align} & 2{{\sin }^{2}}x+3\sin x+1=0 \\ & \left( 2\sin x-1 \right)\left( \sin x-1 \right)=0 \end{align} It implies. $2\sin x-1=0$ Or $sinx-1=0$ Now, \begin{align} & 2\sin x-1=0 \\ & \sin x=\frac{1}{2} \\ & x=\frac{\pi }{6},\frac{5\pi }{6} \end{align} And, \begin{align} & sinx-1=0 \\ & \sin x=1 \\ & x=\frac{\pi }{2} \end{align} Thus, the solutions are $\frac{\pi }{2},\frac{\pi }{6},\frac{5}{6}$.