Answer
$\dfrac{5\pi}{6}$
Work Step by Step
We need to find: $\cos^{-1} (-\dfrac{\sqrt 3}{2})$
The inverse cosine function has the range $0\le \theta \le \pi$. We are looking for the angle whole cosine value equals zero.
We know from the unit circle that
$\cos (\dfrac{5 \pi}{6})=\sin (\dfrac{\pi}{2}-\dfrac{5 \pi}{6}) =\sin (-\dfrac{\pi}{3}) =-\dfrac{\sqrt 3}{2}$
Since $\dfrac{ 5\pi}{6}$ lies in $0\le \theta \le \pi$, then we have:
$\cos^{-1} (-\dfrac{\sqrt 3}{2})=\dfrac{5\pi}{6}$
