Answer
$ 240^{\circ}$ and $300^{\circ}$
Work Step by Step
$\sin\theta=-\frac{\sqrt 3}{2}$
The value of $\sin \theta$ is negative, so $\theta$ may lie in either quadrant III or IV.
We know that $\sin 60^{\circ}=\frac{\sqrt 3}{2}$.
Recall: $-\sin x= \sin(180^{\circ}+x)$ and $-\sin x= \sin(360^{\circ}-x)$
$\implies -\sin 60^{\circ}=-\frac{\sqrt 3}{2}=\sin(180^{\circ}+60^{\circ})$
$=\sin 240^{\circ}$
$240^{\circ}$ which is in the third quadrant is one value of $\theta$.
Now $-\sin 60^{\circ}=-\frac{\sqrt 3}{2}= \sin (360^{\circ}-60^{\circ})$
$=\sin 300^{\circ}$
$300^{\circ}$ is in the fourth quadrant.
The values of $\theta$ are $ 240^{\circ}$ and $300^{\circ}$.
