Answer
$\theta=-60^o$
Work Step by Step
RECALL:
$y=\arcsin{(x)} \longrightarrow \sin{y}=x$, $y$ is in the interval $[90^o, 90^o]$
Thus, $\theta=\arccos{(-\frac{\sqrt3}{2})}$ implies that $\cos{\theta}=-\frac{\sqrt3}{2}$.
Note that $\sin{(60^o)}=\frac{\sqrt3}{2}$.
Since sine is an odd function, then $\sin{(-x)}=-\sin{x}$.
This means that $\sin{(-60^o)}=-\sin{(60^o)}=-\frac{\sqrt3}{2}$
Therefore,
$\theta=\arcsin{(-\frac{\sqrt3}{2})}\longrightarrow \theta=-60^o$
