Question

solve the question tan X + tan 2 X + tan X tan 2x =1

Answer 1

tan x+tan2x+tanx tan2x=1

Answer 2

Recall the identity,

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$.

Now,

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

Note that $\cos 3x \neq 0$

$\tan 2x+\tan x+\tan 2x\tan x-1=(1-\tan 2x \tan x)\tan 3x+\tan 2x\tan x-1=0\\ (1-\tan 2x \tan x)(\tan 3x-1)=0\\ (1-\tan 2x \tan x)(\tan 3x-1)=0\\ (1-\frac{\sin 2x \sin x}{\cos 2x \cos x})(\tan 3x-1)=0\\ (\frac{\cos 2x \cos x-\sin 2x \sin x}{\cos 2x \cos x})(\tan 3x-1)=0\\ (\frac{\cos 3x }{\cos 2x \cos x})(\tan 3x-1)=0\\ (\frac{\cos 3x }{\cos 2x \cos x})(\tan 3x-1)=0$

The solution of the above equation are,

$\cos 3x=0,\tan 3x=1$

The general solution of $\cos 3x=0$ is

$3x=2n\pi \pm \frac{\pi}{2} ,n=0,1,2,3,...\\ x=\frac{2n\pi}{3} \pm \frac{\pi}{6} ,n=0,1,2,3,...$

This is ignored since $\cos 3x \neq 0$.

The general solution of $\tan 3x=1$ is

$3x=n\pi + \frac{\pi}{4} ,n=0,1,2,3,...\\ x=\frac{n\pi }{3}+ \frac{\pi}{12} ,n=0,1,2,3,...$

To sum up the general solution of the given equation,

$x=\frac{n\pi }{3}+ \frac{\pi}{12},n=0,1,2,3,...$