Question

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

Answer 1

Solve for principal solution.
Given $\cot x=-\sqrt{3}\Rightarrow \frac{1}{\tan x}=-\sqrt{3}\Rightarrow \tan x=-\frac{1}{\sqrt{3}}$

We know that $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$
Since $\tan x$ is negative,
So, $\tan x$ will lie in $2^{nd}$ and $4^{th}$ quadrant.

Value in $2^{nd}$ quadrant. $={180}^{\circ }-{30}^{\circ }={150}^{\circ }$
Value in $4^{th}$ quadrant. $={360}^{\circ }-{30}^{\circ }={330}^{\circ }$
So, principal solutions are $x={150}^{\circ }=150\times \frac{\pi }{180}=\frac{5\pi }{6}$ and $x={330}^{\circ }=330\times \frac{\pi }{180}=\frac{11\pi }{6}$

Step 2: Solve for general solution
$\tan x=-\frac{1}{\sqrt{3}}$
$\Rightarrow \tan x=\tan \frac{5\pi }{6}$

We know that if $\tan x=\tan y$
General solution is $x=n\pi +y,$ where $n\in Z$
Put $y=\frac{5\pi }{6}$
Hence, $x=n\pi +\frac{5\pi }{6},$ where $n\in Z$