Question
Find the principal and general solution of the equation: cot x = - $\sqrt 3$

Answer

Given: cot x = $-\sqrt{3}$
We know that, $\cot \frac{\pi}{6}=\sqrt{3} \text { and } \cot \left(\pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$
Also $\cot \left(2 \pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$
$\Rightarrow \cot \frac{5 \pi}{6}=-\sqrt{3} \text { and } \cot \frac{11 \pi}{6}=-\sqrt{3}$
$\therefore$ the principal solutions of the equat̥ion are $x=\frac{5 \pi}{6} \text { and } \frac{11 \pi}{6}$
Now when, $\cot x=\cot \frac{5 \pi}{6}$
$\Rightarrow \tan x=\tan \frac{5 \pi}{6}\left(\because \cot x=\frac{1}{\tan x}\right)$
$\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{5 \pi}{6}$, where n$\in$ z and z is set of integers.
Hence the general solution of the equation $\mathrm{x}=\mathrm{n} \pi+\frac{5 \pi}{6}$ where n $\in$ Z and z is set of integers.

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