Answer
$\dfrac{\pi}{6}$
Work Step by Step
We are given the expression: $\cos^{-1} \dfrac{\sqrt 3}{2}$
$y=\cos^{-1} x$ is the value of $y$ so that $\cos y=x$, where $0\leq y\leq\pi$.
So $y=\cos^{-1} \dfrac{\sqrt 3}{2}$ is the value of $y$ so that $\cos y=\dfrac{\sqrt 3}{2}$, where $0\leq y\leq\pi$
From the unit circle we have:
$x=\cos \left(\dfrac{\pi}{6}\right)$
Therefore the value of $y$ for which $x=\dfrac{\sqrt 3}{2}$ in the interval $\left[0,\pi\right]$ is $\dfrac{\pi}{6}$, so we got:
$\cos^{-1} \dfrac{\sqrt 3}{2}=\dfrac{\pi}{6}$
