Answer
The solutions are $\frac{\pi }{2},\frac{\pi }{6},\frac{5}{6}$.
Work Step by Step
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 $\frac{\pi }{2},\frac{\pi }{6},\frac{5}{6}$.
