Question

Find principal and general solution of the equation, $\cot x=-\sqrt{3}$.

Answer 1

$\cot x=-\sqrt{3}$

We know, $\cot \frac{\pi }{6}=\sqrt{3}$

Therefore, $\cot (\pi -\frac{\pi }{6})=-\cot \frac{\pi }{6}=-\sqrt{3}$ and $\cot (2\pi -\frac{\pi }{6})=-\cot \frac{\pi }{6}=-\sqrt{3}$

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

Therefore, the principal solutions are $x=\frac{5\pi }{6}$ and $\frac{11\pi }{6}$.

Now, $\cot x=\cot \frac{5\pi }{6}$

$\tan x=\tan \frac{5\pi }{6}$

$x=n\pi +\frac{5\pi }{6}$

Therefore, the general solution is $x=n\pi +\frac{5\pi }{6},n\in Z$.