Question

General solution of the equation $\tan x+\tan 2x+\sqrt{3}\tan x\tan 2x=\sqrt{3}$ is :

• A. $\frac{n\pi }{3}+\frac{\pi }{9},n\in Z$
• B. $n\pi +\frac{\pi }{9},n\in Z$
• C. $n\pi -\frac{\pi }{9},n\in Z$
• D. $\frac{n\pi }{2}+\frac{\pi }{3},n\in Z$

Answer 1

The correct option is A $\frac{n\pi }{3}+\frac{\pi }{9},n\in Z$

$\tan x+\tan 2x+\sqrt{3}\tan x\tan 2x=\sqrt{3}\Rightarrow \tan x+\tan 2x=\sqrt{3}(1-\tan x\tan 2x)\Rightarrow \frac{\tan x+\tan 2x}{(1-\tan x\tan 2x)}=\sqrt{3}\tan (2x+x)=\sqrt{3}\Rightarrow \tan 3x=\sqrt{3}\Rightarrow \tan 3x=\tan \frac{\pi }{3}\Rightarrow 3x=n\pi +\frac{\pi }{3}\Rightarrow x=\frac{n\pi }{3}+\frac{\pi }{9},n\in Z$