Answer
$\dfrac{\pi}{3}$
Work Step by Step
We are given the expression: $\sin^{-1} \dfrac{\sqrt 3}{2}$
$y=\sin^{-1} x$ is the value of $y$ so that $\sin y=x$, where $-\dfrac{\pi}{2}\leq y\leq\dfrac{\pi}{2}$.
So $y=\sin^{-1} \dfrac{\sqrt 3}{2}$ is the value of $y$ so that $\sin y=\dfrac{\sqrt 3}{2}$, where $-\dfrac{\pi}{2}\leq y\leq\dfrac{\pi}{2}$
From the unit circle we have:
$x=\sin \left(\dfrac{\pi}{3}\right)$
Therefore the value of $y$ for which $x=\dfrac{\sqrt 3}{2}$ in the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ is $\dfrac{\pi}{3}$, so we got:
$\sin^{-1} \dfrac{\sqrt 3}{2}=\dfrac{\pi}{3}$
