Answer
$\frac{7\pi}{6}$, $\frac{3\pi}{2}$, $\frac{11\pi}{6}$
Work Step by Step
$-2\sin^2x=3\sin x+1$
$2\sin^2x+3\sin x+1=0$
$(2\sin x+1)(\sin x+1)=0$
$2\sin x+1=0$ or $\sin x+1=0$
If $2\sin x+1=0$:
$2\sin x=-1$
$\sin x=-\frac{1}{2}$
The only solutions in $[0, 2\pi)$ are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.
If $\sin x+1=0$:
$\sin x=-1$
The only solution in $[0, 2\pi)$ is $\frac{3\pi}{2}$.
