## Precalculus (6th Edition) Blitzer

$$x=\frac{3\pi }{2}$$
We solve by factoring: $$2\sin ^2x - \sin x -3=0 \quad \Rightarrow \quad (\sin x + 1)(\sin x - \frac{3}{2})=0 \quad \Rightarrow \quad \sin x +1=0 \quad \Rightarrow \quad \sin x= -1 \quad \Rightarrow \quad x= \frac{3\pi }{2}+2n \pi, \quad n \in \mathbb {Z}$$Please note that the range of $\sin x$ is $[-1,1]$. So, $x=\frac{3\pi }{2}$ is the only solution in the interval $[0, 2\pi)$.