Question

Let $S=\{x\in (-\pi ,\pi ):x\ne 0,\pm \frac{\pi }{2}\}$. The sum of all distinct solutions $\sqrt{3}\sec x+\csc x+2\left(\tan x-\cot x\right)=0$ in the set S is equal to

• A. $-\frac{7\pi }{9}$
• B. $-\frac{2\pi }{9}$
• C. $0$
• D. $\frac{5\pi }{9}$

Answer 1

The correct option is C $0$

$\sqrt{3}\sec x+\csc x=2(\cot x-\tan x)$

$\Rightarrow \frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}=2[\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}]$

$\sqrt{3}\sin x+\cos x=2({\cos }^{2}x-{\sin }^{2}x)$

Dividing both sides by 2,

$\Rightarrow \frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x=\cos 2x$

$\Rightarrow \cos {60}^{\circ }\sin x+\cos {60}^{\circ }\cos x=\cos 2x$

$\Rightarrow \cos (x-\frac{\pi }{3})=\cos 2x$

$\therefore 2x=2n\pi \pm (x-\frac{\pi }{3})$

When $2x=2n\pi +x-\frac{\pi }{3}$

$\Rightarrow x=2n\pi -\frac{\pi }{3}$

When $2x=2n\pi -x+\frac{\pi }{3}$

$\Rightarrow 3x=2n\pi +\frac{\pi }{3}$

$\Rightarrow x=\frac{2n\pi }{3}+\frac{\pi }{9}$

For $n=0,x=\frac{-\pi }{3}x=\frac{\pi }{9}$

For $n=-1,x=\frac{-7\pi }{3}x=\frac{-2\pi }{3}+\frac{\pi }{9}=\frac{-5\pi }{9}$

For $n=1,x=\frac{5\pi }{3}x=\frac{2\pi }{3}+\frac{\pi }{3}=\frac{7\pi }{3}$

$\therefore x=\frac{-\pi }{3},\frac{\pi }{9},\frac{-5\pi }{9},\frac{7\pi }{9}$

$\therefore$ Sum of distinct solutions$=\frac{-\pi }{3}+\frac{\pi }{9}+\frac{-5\pi }{9}+\frac{7\pi }{9}$

$=0$