Answer
$$\frac{\pi}{3}.$$
Work Step by Step
Since $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\frac{\pi}{3}\in[-\frac{\pi}{2},\frac{\pi}{2}]$, we have
$$\sin^{-1}\left(\sin\frac{\pi}{3}\right)=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}.$$
