Question

Number of solutions of the equation $2+ta{n}^{2}x-2tanx-sin2x=0$, where $x\in [-\pi ,3\pi ]$ is.

Answer 1

$2+{\tan }^{2}x-2\tan x-\sin 2x=0$

$\Rightarrow {\tan }^{2}x-2\tan x+1+1-\sin 2x=0$

$\Rightarrow {(\tan x-1)}^2+{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=0$ using $1={\sin }^{2}x+{\cos }^{2}x$

$\Rightarrow {(\tan x-1)}^2+{(\sin x-\cos x)}^2=0$

$\Rightarrow {(\tan x-1)}^2=0$ and ${(\sin x-\cos x)}^2=0$

$\Rightarrow \tan x=1$ and $\sin x=\cos x$

$\Rightarrow \tan x=1$

$\therefore x=n\pi +\frac{\pi }{4}$

For $n=0\Rightarrow x=\frac{\pi }{4}$

For $n=-1\Rightarrow x=-\pi +\frac{\pi }{4}=\frac{-3\pi }{4}$

For $n=1\Rightarrow x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}$

For $n=2\Rightarrow x=2\pi +\frac{\pi }{4}=\frac{9\pi }{4}$

For $x\in [-\pi ,3\pi ]$

Number of solutions$=\{\frac{-3\pi }{4},\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}\}=4$