Answer
$$\theta = \frac{\pi }{2} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{2} + 2n\pi ,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots $$
Work Step by Step
$$\eqalign{
& {\sin ^2}\theta - 1 = 0 \cr
& {\text{Add 1 to both sides of the equation}} \cr
& {\sin ^2}\theta = 1 \cr
& \sin \theta = \pm 1 \cr
& \sin \theta = 1{\text{ or sin}}\theta = - 1 \cr
& {\text{In the interval }}\left[ {0,2\pi } \right)\,\,\sin \theta = 1{\text{ for }}\theta = \frac{\pi }{2}{\text{ and sin}}\theta = - 1{\text{ for}} \cr
& \theta = \frac{{3\pi }}{2} \cr
& {\text{Moreover, because }}\sin \theta {\text{ has a period of 2}}\pi ,{\text{ there are infinitely }} \cr
& {\text{many other solutions, which can be written as}} \cr
& \theta = \frac{\pi }{2} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{2} + 2n\pi \cr
& {\text{Where }}n{\text{ is an integer}}{\text{.}} \cr
& {\text{The general solution is:}} \cr
& \theta = \frac{\pi }{2} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{2} + 2n\pi ,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots \cr} $$
