Solution:

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right)$
[ Formula: $\sin (\pi-x)=\sin x)$
$=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)$
[ Formula: $\left.\sin ^{-1}(\sin x)=x\right]$
$=\frac{\pi}{3}$