Question

The principle solution of equation $\cot x=-\sqrt{3}$ is

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{5\pi }{6}$

Answer 1

The correct option is D $\frac{5\pi }{6}$

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

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

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

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

We know that $\tan {30}^o=1/\sqrt{3}$

We find the value of x where $\tan$ is negative

$\tan$ is negative in $2$ and $4^{th}$ quadrant.

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

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

So, principal solution are

$x={150}^o$ and $x={330}^o$

$x=150\times \pi /180$ and $x=330\times \pi /180$

$x=5\pi /6$ and $x=11\pi /6$

To find general solution

Let $\tan x=\tan y$ ………..$\left(1\right)$

$\tan x=-1/\sqrt{3}$ ………..$\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$\tan y=-1/\sqrt{3}$

$\tan y=\tan 5\pi /6$

$y=5\pi /6$

Since $\tan x=\tan y$

General solution is $x=n\pi +y$ where $n\in z$.

Hence $x=n\pi +5\pi /6$ where $n\in z$.