Answer
$$0, \frac{\pi}{3}, \pi, \frac{5 \pi}{3}$$
Work Step by Step
Since
\begin{aligned}
\sin \theta&=\sin 2 \theta\\
\sin \theta&=2 \sin \theta \cos \theta \\
\sin \theta-2 \sin \theta \cos \theta&=0\\
\sin \theta(1-2 \cos \theta)&=0
\end{aligned}
Where we used the double angle identity
$\sin 2 \theta=2 \sin \theta \cos \theta$
Then
$$ \sin \theta=0 \text { or } 1-2 \cos \theta=0$$
Hence
$$\theta \in\left\{0, \frac{\pi}{3}, \pi, \frac{5 \pi}{3}\right\}$$
