# How to solve?

## Solve $\tan x + \tan 2 x + \tan 3 x = 0$ without changing tan into sin and cos.

Jan 24, 2018

The Solution Set is given by,

$\left\{k \frac{\pi}{3}\right\} \cup \left\{k \pi \pm a r c \tan \left(\frac{1}{\sqrt{2}}\right)\right\} , k \in \mathbb{Z}$.

#### Explanation:

Let, $\tan x = u , \tan 2 x = v$.

Then, $\tan 3 x = \tan \left(x + 2 x\right) = \frac{\tan x + \tan 2 x}{1 - \tan x \tan 2 x} , i . e . ,$

$\tan 3 x = \frac{u + v}{1 - u v}$.

$\therefore \tan x + \tan 2 x + \tan 3 x = 0 ,$

$\Rightarrow u + v + \frac{u + v}{1 - u v} = 0$,

$\Rightarrow \left(u + v\right) \left\{1 + \frac{1}{1 - u v}\right\} = 0$,

$\Rightarrow \frac{\left(u + v\right) \left(2 - u v\right)}{1 - u v} = 0$,

$\Rightarrow \left(u + v\right) \left(2 - u v\right) = 0$,

$\Rightarrow v = - u , \mathmr{and} , u v = 2$.

$v = - u \Rightarrow \tan 2 x = - \tan x = \tan \left(- x\right)$.

Recall that, $\tan \theta = \tan \alpha \Rightarrow \theta = k \pi + \alpha , k \in \mathbb{Z}$.

$\text{Case 1 : } v = - u$

$\therefore \tan 2 x = \tan \left(- x\right)$

$\Rightarrow 2 x = k \pi + \left(- x\right) , \mathmr{and} , 3 x = k \pi , i . e . , x = k \frac{\pi}{3} , k \in \mathbb{Z}$.

$\text{Case 2 : } u v = 2$.

$\therefore \tan x \cdot \tan 2 x = 2$,

Using $\tan 2 x = \frac{2 \tan x}{1 - {\tan}^{2} x}$, we have, then

$\frac{2 {\tan}^{2} x}{1 - {\tan}^{2} x} = 2 , \mathmr{and} , {\tan}^{2} x = 1 - {\tan}^{2} x , i . e . ,$

${\tan}^{2} x = \frac{1}{2} \Rightarrow \tan x = \pm \frac{1}{\sqrt{2}}$,

$\Rightarrow x = k \pi + a r c \tan \left(\pm \frac{1}{\sqrt{2}}\right) , k \in \mathbb{Z}$.

Altogether, the Solution Set is given by,

$\left\{k \frac{\pi}{3}\right\} \cup \left\{k \pi \pm a r c \tan \left(\frac{1}{\sqrt{2}}\right)\right\} , k \in \mathbb{Z}$.