# How do you evaluate Arccos (- sqrt 3 / 2)?

$\frac{5 \pi}{6} \mathmr{and} \frac{7 \pi}{6}$
Trig table and unit circle --> $\cos x = \left(- \frac{\sqrt{3}}{2}\right)$ -->
arc $x = \pm \frac{5 \pi}{6}$
Answers for $\left(0 , 2 \pi\right)$;
$\frac{5 \pi}{6} \mathmr{and} \frac{7 \pi}{6}$ (co-terminal to $- \frac{5 \pi}{6}$)