Question

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

Answer 1

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

$\tan x=\frac{1}{\cot x}$

$\tan x=\frac{1}{-\sqrt{3}}$

We know that $\tan 30=\frac{1}{\sqrt{3}}$

We find the value of x where tan is negative

tan is negative in 2nd and 4th quadrant

Value in 2nd quadrant $=180-30={150}^o$

Value in 4th quadrant $=360-30={330}^o$

So. principal solution are

$x={150}^o$

$x=150\times \frac{\pi }{180}$

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

$x={330}^o$

$x=330\times \frac{\pi }{180}$

$x=\frac{11\pi }{6}$