Find the principal solutions of the equation $\tan x=-\frac{1}{\sqrt{3}}.$
We know that, $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} .$
Thus, $\tan \left(\pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$
and $\quad \tan \left(2 \pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$
Thus $\quad \tan \frac{5 \pi}{6}=\tan \frac{11 \pi}{6}=-\frac{1}{\sqrt{3}}$
Therefore, principal solutions are $\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$ .
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The positive integer value of $n>3$ satisfying the equation $\frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)}$ is
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