Answer
$$x=\frac{\pi }{2}, \qquad x=\frac{3\pi }{2}$$
Work Step by Step
We solve by factoring:
$$\sin ^2 \theta -1=0 \quad \Rightarrow \quad \sin ^2 \theta =1 \quad \Rightarrow \quad \sin \theta = \pm 1 \quad \Rightarrow \quad x=\frac{\pi }{2}+ 2n\pi, \quad \text{ or } \quad x=\frac{3\pi }{2}+ 2n \pi, \quad n \in \mathbb{Z}$$So, $x= \frac{\pi }{2}$ and $x= \frac{3\pi }{2}$ are the only solutions in the interval $[0, 2\pi )$.
