# How do you calculate arccos(- sqrt3/2)?

Since the range of the function $y = \arccos x$ is $\left[0 , \pi\right]$ and the value negative $- \frac{\sqrt{3}}{2}$, the angle is in the second quadrant.
$\arccos \left(- \frac{\sqrt{3}}{2}\right) = \frac{5}{6} \pi$.