Question

The general solution of the equation
$tanx+tan2x+\sqrt{3}tanxtan2x=\sqrt{3}$ is

• A. $x=\frac{n\pi }{3}+\frac{\pi }{9},n\in z$
• B. $x=m\pi +\frac{\pi }{9},n\in z$
• C. $x=\frac{(n+1)\pi }{3},n\in z$
• D. $x=\frac{n\pi }{3}+\frac{\pi }{3},n\in z$

Answer 1

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

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

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

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

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

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

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

$\Rightarrow 3x=n\pi +\frac{\pi }{3}$ ( where $n$ is an integer )

$x=\frac{n\pi }{3}+\frac{\pi }{9}$