# How do you solve 3 tan 2 x = √ 3 tan x in interval 0<=x<=360?

May 2, 2018

$x = \left\{{0}^{\circ} , {180}^{\circ} , {360}^{\circ}\right\}$

#### Explanation:

As $\tan 2 x = \frac{2 \tan x}{1 - {\tan}^{2} x}$, we can write $3 \tan 2 x = \sqrt{3} \tan x$ as

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

or $6 \tan x = \sqrt{3} \tan x - \sqrt{3} {\tan}^{3} x$

or $\sqrt{3} {\tan}^{3} x + \tan x \left(6 - \sqrt{3}\right) = 0$

or $\sqrt{3} \tan x \left({\tan}^{2} x + 2 \sqrt{3} - 1\right) = 0$

As ${\tan}^{2} x + 2 \sqrt{3} - 1 \ne 0$,

hence $\tan x = 0$ i.e. $x = n \pi$, where $n$ is an integer

and in ${0}^{\circ} \le x \le {360}^{\circ}$, $x = \left\{{0}^{\circ} , {180}^{\circ} , {360}^{\circ}\right\}$.

May 2, 2018

$\implies x = {0}^{\circ} , {180}^{\circ} , {360}^{\circ} .$

#### Explanation:

Here,

3 tan 2 x = √ 3 tan x

$\implies 3 \left(\frac{2 \tan x}{1 - {\tan}^{2} x}\right) = \sqrt{3} \tan x$

$\implies 6 \tan x = \sqrt{3} \tan x - \sqrt{3} {\tan}^{3} x$

$\implies \sqrt{3} {\tan}^{3} x + 6 \tan x - \sqrt{3} \tan x = 0$

$\implies \tan x \left(\sqrt{3} {\tan}^{2} x + 6 - \sqrt{3}\right) = 0$

$\implies \tan x = 0 \mathmr{and} \sqrt{3} {\tan}^{2} x + 6 - \sqrt{3} = 0$

$\left(i\right) \tan x = 0 \implies x = k \pi , k \in \mathbb{Z}$

$\left(i i\right) \sqrt{3} {\tan}^{2} x + 6 - \sqrt{3} = 0$

$\implies {\tan}^{2} x = - \frac{6 - \sqrt{3}}{\sqrt{3}} = - \left(2 \sqrt{3} - 1\right) < 0$

$i . e . {\tan}^{2} x < 0 \implies x \notin \mathbb{R}$

Hence,

$x = k \pi , k \in \mathbb{Z} , w h e r e , x \in \left[{0}^{\circ} , {360}^{\circ}\right]$

$\implies x = {0}^{\circ} , {180}^{\circ} , {360}^{\circ} .$