Question

The general solution of $\cos 2x-2\tan x+2=0$ is

• A. $(2n+1)\frac{\pi }{3},n\in Z$
• B. $(n+1)\frac{\pi }{3},n\in Z$
• C. $n\pi +\frac{\pi }{3},n\in Z$
• D. $n\pi +\frac{\pi }{4},n\in Z$

Answer 1

Answer: (D)

We have,
$\cos 2x-2\tan x+2=0$
$\Rightarrow \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}-2\tan x+2=0$
$\Rightarrow \left(1-{\tan }^{2}x\right)-2(\tan x-1)\left(1+{\tan }^{2}x\right)=0$
$\Rightarrow (1-\tan x)\left(1+\tan x+2+2{\tan }^{2}x\right)=0$
$\Rightarrow (1-\tan x)\left(2{\tan }^{2}x+\tan x+3\right)=0$
$\Rightarrow \tan x=1\Rightarrow \tan x=\tan \frac{\pi }{4}$
$\Rightarrow x=n\pi +\frac{\pi }{4}$