Question

The general solution of $\frac{\tan 2x-\tan x}{1+\tan x\tan 2x}=1$ is

• A. $n\pi +\frac{\pi }{4}$
• B. $n\pi \pm \frac{\pi }{3}$
• C. $\varphi$
• D. $n\pi +\frac{\pi }{6}$

Answer 1

The correct option is C $\varphi$

$\frac{\tan 2x-\tan x}{1+\tan x\tan 2x}=1\Rightarrow \tan (2x-x)=1\Rightarrow \tan x=\tan (n\pi +\frac{\pi }{4})\Rightarrow x=(n\pi +\frac{\pi }{4})$
But $\tan 2x$ is not defined at $(n\pi +\frac{\pi }{4})$, so this trigonometric equation has no solution.