Question

The number of solutions of the equation $\sqrt{13-18\tan x}=6\tan x-3$, where $-2\pi <x<2\pi$ is

• A. 0
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (C)

$\sqrt{13-18\tan x}=6\tan x-3-2\pi <x<2\pi$ (given)
$\Rightarrow 13-18\tan x=36{\tan }^{2}x+9-36\tan x$
$\Rightarrow 36{\tan }^{2}x-18\tan x-4=0$
$\Rightarrow 36{\tan }^{2}x-24\tan x+6\tan x-4=0$
$\Rightarrow 6\tan x(6\tan x+1)-4(6\tan x+1)=0$
$\tan x=\frac{2}{3}$
or $\tan x=-1/6$ (not possible)
Let $\tan \alpha =2/3$
then $\tan x=\tan \alpha$
or $x=\tan (2/3),\pi +{\tan }^{-1}(2/3),{\tan }^{-1}(2/3)-\pi ,{\tan }^{-1}(2/3)-2\pi$