Find the solution of $\sin x=-\frac{\sqrt{3}}{2}$
Find the solution of $\sin x=-\frac{\sqrt{3}}{2}$
We have $\sin x=-\frac{\sqrt{3}}{2}$
$=-\sin \frac{\pi}{3}=\sin \left(\pi+\frac{\pi}{3}\right)$
$=\sin \frac{4 \pi}{3}$
Hence $\sin x=\sin \frac{4 \pi}{3},$ which gives
$x=n \pi+(-1)^{n} \frac{4 \pi}{3}, \text { where } n \in Z$
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