Question
Find the principal solutions of $\cot \theta=-\sqrt{3}$

Answer

We know that $\cot \theta=-\sqrt{3}$ if and only if $\tan \theta=-\frac{1}{\sqrt{3}}$

We know that $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$

Using identities, $\tan (\pi-\theta)=-\tan \theta$ and $\tan (2 \pi-\theta)=-\tan \theta$, we get

$\tan \frac{5 \pi}{6}=-\frac{1}{\sqrt{3}}$ and $\tan \frac{11 \pi}{6}=-\frac{1}{\sqrt{3}}$

An $0 \leq \frac{5 \pi}{6}<2 \pi$ and $0 \leq \frac{11 \pi}{6}<2 \pi$

$\therefore \frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$ are required principal solutions.

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