Question

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

• A. $\frac{n\pi }{3}+\frac{\pi }{9}$
• B. $\frac{n\pi }{3}+\frac{2\pi }{9}$
• C. $\frac{2n\pi }{3}+\frac{\pi }{9}$
• D. None of these

Answer 1

Answer: (A)

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

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

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

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

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

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

$\Rightarrow \tan (x+2x)=\tan \left(\pi /3\right)$

$\Rightarrow \tan 3x=\tan \left(\pi /3\right)$

$\Rightarrow 3x=n\pi +\pi /3$ $[\because \tan \theta =\tan \alpha \theta =n\pi +\alpha \forall n\in z]$

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

$x=\frac{n\pi }{3}+\frac{\pi }{9},\forall n\in z$