Question

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

• A. General solution $=n\pi +\frac{5\pi }{3}$ and Principal solution $=\frac{5\pi }{3},\frac{11\pi }{3}$
• B. General solution $=n\pi -\frac{5\pi }{3}$ and Principal solution $=\frac{5\pi }{3},\frac{11\pi }{3}$
• C. General solution $=n\pi -\frac{5\pi }{6}$ and Principal solution $=\frac{5\pi }{6},\frac{11\pi }{6}$
• D. General solution $=n\pi +\frac{5\pi }{6}$ and Principal solution $=\frac{5\pi }{6},\frac{11\pi }{6}$

Answer 1

Answer: (D)

General solution $=n\pi +\frac{5\pi }{6}$ and Principal solution $=\frac{5\pi }{6},\frac{11\pi }{6}$

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

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

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

We know that: $-\tan {30}^o=\frac{-1}{\sqrt{3}}$

We find value of $x$ where $x$ is negative. $\tan$ is negative in $II$ and $IV$ quadrant.

So, principal solutions are

$x=\frac{5\pi }{6},x=\frac{11\pi }{6}$

To find general solution:

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

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

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

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

$\tan y=\tan \frac{5\pi }{6}$

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

Since $\tan x=\tan y$

General solution is $x=n\pi +y,n\in Z$

$x=n\pi +\frac{5\pi }{6},n\in Z$