cot x = − $\sqrt{3}$ ⇒ tan x = $-\frac{1}{\sqrt{3}}$ A value of x satisfying tanx = $\frac{1}{\sqrt{3}}$ is $\frac{\pi}{6}$ Principal Solution : We have tanx = $-\frac{1}{\sqrt{3}}$ Thus , tan $\left(\pi - \frac{\pi}{6}\right)$ = - tan $\frac{\pi}{6}$ = $\frac{-1}{\sqrt{3}}$ and , tan $\left(2\pi - \frac{\pi}{6}\right)$ = - tan $\frac{\pi}{6}$ = - $\frac{1}{\sqrt{3}}$ Thus , tan $\frac{5\pi}{6}$ = tan $\frac{11\pi}{6}$ = $\frac{-1}{\sqrt{3}}$ Therefore principal solutions are $\frac{5\pi}{6},\frac{11\pi}{6}$ General Solution : We have, tanx = $\frac{-1}{\sqrt{3}}$ ⇒ tan x = tan $\left(\pi - \frac{\pi}{6}\right)$ = tan $\frac{5\pi}{6}$ ⇒ x = nπ + $\frac{5\pi}{6}$ , n ∊ Z Since tan x = tan α , then x = nπ + α , n ∊ Z