Question

Solve: $\tan x+\tan 2x+\sqrt{3}\tan x\tan 2x=\sqrt{3}$

Answer 1

$\tan x+\tan 2x+\sqrt{3}\tan x\tan 2x=\sqrt{3}$

$\Rightarrow \tan x+\tan 2x=\sqrt{3}-\sqrt{3}\tan x\tan 2x$

$\Rightarrow \tan x+\tan 2x=\sqrt{3}(1-\tan x\tan 2x)$

$\Rightarrow \frac{\tan x+\tan 2x}{1-\tan x\tan 2x}=\sqrt{3}=\tan \frac{\pi }{3}$

$\Rightarrow \tan 3x=\tan \frac{\pi }{3}$

$\Rightarrow 3x=n\pi +\frac{\pi }{3}$

$\Rightarrow x=\frac{n\pi }{3}+\frac{\pi }{9}$ where $n\in Z$

$\therefore x=\{\frac{\pi }{9},\frac{\pi }{3}+\frac{\pi }{9},\frac{2\pi }{3}+\frac{\pi }{9},\pi +\frac{\pi }{9},\frac{4\pi }{3}+\frac{\pi }{9},\frac{5\pi }{3}+\frac{\pi }{9}\}$

or $x=\{\frac{\pi }{9},\frac{4\pi }{9},\frac{7\pi }{9},\frac{10\pi }{9},\frac{13\pi }{9},\frac{16\pi }{9}\}$