Question

Find the principal and general solutions of the following equations.
cot x =$-\sqrt{3}$

Answer 1

$\text{Here cot x =}$$-\sqrt{3}\Rightarrow tanx=-\frac{1}{\sqrt{3}}$, $\text{which is negative, so x lies in second or fourth quadrant.}$

tan x = -$\frac{1}{\sqrt{3}}$ = - tan ${30}^0$

= tan $({180}^0-{30}^0)$ or tan $({360}^0-{30}^0)$

= tan ${150}^0$ or tan ${330}^0$

= tan $\frac{5\pi }{6}$ or tan $\frac{11\pi }{6}$

Here the prinicipal solutions are $\frac{5\pi }{6},\frac{11\pi }{6}$.

Now tan x =- tan $\frac{\pi }{6}$

$\Rightarrow x=n\pi +\frac{5}{6}\pi$ where n $ϵ$ Z.