Question

The general solution of $\tan x+\tan 2x+\sqrt{3}\tan x\cdot \tan 2x=\sqrt{3}$ is
$(\text{where}n\in Z)$

• A. $\frac{n\pi }{3}+(-1{)}^n\frac{\pi }{6}$
• B. $2n\pi \pm \frac{\pi }{6}$
• C. $\frac{n\pi }{9}+\frac{\pi }{3}$
• D. $\frac{n\pi }{3}+\frac{\pi }{9}$

Answer 1

The correct option is D $\frac{n\pi }{3}+\frac{\pi }{9}$

Given:
$\tan x+\tan 2x+\sqrt{3}\tan x\cdot \tan 2x=\sqrt{3}\Rightarrow \tan x+\tan 2x=\sqrt{3}(1-\tan x\cdot \tan 2x)\Rightarrow \frac{\tan x+\tan 2x}{1-\tan x\cdot \tan 2x}=\sqrt{3}\Rightarrow \tan 3x=\tan \frac{\pi }{3}$
So,general solution is $3x=n\pi +\frac{\pi }{3}\Rightarrow x=\frac{n\pi }{3}+\frac{\pi }{9},n\in Z$