Question

Solve for x, $\sqrt{13-18\tan x}=6\tan x-3$, where $-2\pi <x<2\pi$.

• A. $\alpha -2\pi$, where $\alpha ={\tan }^{-1}\frac{2}{3}$
• B. $\alpha -\pi$, where $\alpha ={\tan }^{-1}\frac{2}{3}$
• C. $\alpha$, where $\alpha ={\tan }^{-1}\frac{2}{3}$
• D. $\alpha +\pi$, where $\alpha ={\tan }^{-1}\frac{2}{3}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\alpha -2\pi$, where $\alpha ={\tan }^{-1}\frac{2}{3}$
B $\alpha -\pi$, where $\alpha ={\tan }^{-1}\frac{2}{3}$
C $\alpha$, where $\alpha ={\tan }^{-1}\frac{2}{3}$
D $\alpha +\pi$, where $\alpha ={\tan }^{-1}\frac{2}{3}$

$\sqrt{13-18\tan x}=6\tan x-3\Rightarrow 13-18\tan x=9-36\tan x+36{\tan }^{2}x\Rightarrow 4+18\tan x-36{\tan }^{2}x=0\Rightarrow (3\tan x-2)(1+6\tan x)=0$
$\Rightarrow \tan x=\frac{2}{3}$ or $\tan x=-\frac{1}{6}$
But $\tan x=-\frac{1}{6}$ does not satisfy equation.
Hence $x=n\pi +{\tan }^{-1}\frac{2}{3}$