Discussion Forum

Let   $S= [ x\epsilon (-\pi,\pi):x\neq0, \pm\frac{\pi}{2}],$ The sum of all distinct solutions of the equation $\sqrt{3}\sec x+ cosec x+2(\tan x-\cot x)=0$  in the set S is equal to 

Answer:

Explanation:

Given,    $\sqrt{3} \sec x+ cosec x+2(\tan x-\cot x)=0,$

$(-\pi <x<\pi)-(0,\pm \pi/2)$

$\Rightarrow$     $\sqrt{3}\sin x+\cos x+2 (\sin^{2}x-\cos^{2}x)=0$

$\Rightarrow$     $ \sqrt{3}\sin x +\cos x-2\cos2x=0$

  Multiplying and dividing by   $\sqrt{a^{2}+b^{2}}.$ i.e

 $\sqrt{3+1}=2$  , we get

 $2(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x)-2\cos2x=0$

=  $(\cos x.\cos\frac{\pi}{3}+\sin x.\sin\frac{\pi}{3})-\cos 2x=0$

  $\cos (x-\frac{\pi}{3})=\cos 2x$

$\therefore$       $\therefore2x=2n\pi\pm (x-\frac{\pi}{3})$

  [since., $\cos\theta=\cos \alpha \Rightarrow\theta=2n\pi\pm \alpha]$

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

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

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

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

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

Now, sum of all distinct solutions

$=-\frac{\pi}{3}+\frac{\pi}{9}-\frac{5\pi}{9}+\frac{7\pi}{9}=0$