Answer
$$ \frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2} $$
Work Step by Step
Since
\begin{align*}
\sin \theta&=\cos 2 \theta\\
&=\cos ^{2} \theta-\sin ^{2} \theta \\
&=1-\sin ^{2} \theta-\sin ^{2} \theta\\
&=1-2 \sin ^{2} \theta
\end{align*}
\begin{align*}
2\sin^2 \theta+\sin \theta-1&=0\\
(\sin \theta+1)(2\sin \theta-1)&=0
\end{align*}
\begin{aligned}
&\text { Now for } \sin \theta+1=0, \text { i.e. } \sin \theta=-1, \text { we get } \theta=\frac{3 \pi}{2}\\
&\text { And for } 2 \sin \theta-1=0, \text { we change it to } 2 \sin \theta=1\\
&\text { So, } \sin \theta=\frac{1}{2} \text { and we get } \theta=\frac{\pi}{6} \text { and } \frac{5 \pi}{6}
\end{aligned}
Then $$\theta \in\left\{\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}\right\}$$
