Question

Solve the equation
$\left(ii\right)\tan x+\tan 2x=\tan 3x$

Answer 1

$\tan x+\tan 2x=\tan 3x$

$\Rightarrow \tan x+\tan 2x=\tan 3x$

$\Rightarrow \tan x+\tan 2x=\tan (2x+x)$

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

$\Rightarrow \tan x+\tan 2x=\left[\frac{\tan 2x+\tan x}{1-\tan 2x\tan x}\right]$

$\Rightarrow \tan x+\tan 2x-\left[\frac{\tan 2x+\tan x}{1-\tan 2x\tan x}\right]=0$

$\Rightarrow (\tan x+\tan 2x)(1-\frac{1}{1-\tan 2x\tan x})=0$

$\Rightarrow (\tan x+\tan 2x)\left(\frac{-\tan 2x\tan x}{1-\tan 2x\tan x}\right)=0$

$\Rightarrow (\tan x+\tan 2x)=0\text{or}\left(\frac{-\tan 2x\tan x}{1-\tan 2x\tan x}\right)=0$

$\Rightarrow (\tan x+\tan 2x)=0\text{or}\tan 2x\tan x=0$

$\Rightarrow (\tan x+\tan 2x)=0\text{or}\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)\tan x=0$

$\Rightarrow (\tan x+\tan 2x)=0\text{or}\left(\frac{2{\tan }^{2}x}{1-{\tan }^{2}x}\right)=0$

$\Rightarrow (\tan x+\tan 2x)=0\text{or}{\tan }^{2}x=0$

$\Rightarrow (\tan x+\tan 2x)=0\text{or}x=n\pi$

Now,

$\tan 2x=-\tan x$

$\Rightarrow \tan 2x=\tan (-x)$

We know the general solution of $\tan x=\tan \alpha$ is
$x=m\pi +\alpha ,m\in Z$

So,
$2x=m\pi -x$

$\Rightarrow 3x=m\pi$

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

Hence, the general solution is
$x=\frac{m\pi }{3}\text{and}x=n\pi ,m,n\in Z$