# How to solve this equation? The topic is Radian Measure of Angle size.

## Solve the equation for 0 ≤ x ≤ 2π.

Jul 25, 2018

$x = \frac{\pi}{3} , \frac{4 \pi}{3} , \frac{3 \pi}{4} , \frac{7 \pi}{4}$

#### Explanation:

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

${\tan}^{2} x + \tan x - \sqrt{3} \tan x - \sqrt{3} = 0$

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

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

$\tan x - \sqrt{3} = 0$ or $\tan x + 1 = 0$

$\tan x - \sqrt{3} = 0$
$\tan x = \sqrt{3}$
$x = \frac{\pi}{3} , \pi + \frac{\pi}{3}$ --> $\tan x$ is positive in the first and third quadrant
$x = \frac{\pi}{3} , \frac{4 \pi}{3}$

$\tan x + 1 = 0$
$\tan x = - 1$
$x = \pi - \frac{\pi}{4} , 2 \pi - \frac{\pi}{4}$ --> $\tan x$ is negative in the second and fourth quadrant
$x = \frac{3 \pi}{4} , \frac{7 \pi}{4}$

Jul 25, 2018

$\frac{\pi}{3} , \frac{3 \pi}{4} , \frac{4 \pi}{3} , \frac{7 \pi}{4}$

#### Explanation:

Given: ${\tan}^{2} x + \tan x = \sqrt{3} \tan x + \sqrt{3} , \text{ in } \left[0 , 2 \pi\right]$

Rearrange the equation to be $= 0$:

${\tan}^{2} x + \tan x - \sqrt{3} \tan x - \sqrt{3} = 0$

Group factor:

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

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

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

tan x = -1; " "tan x = sqrt(3)

The tangent is positive in quadrants I, III and negative in quadrants II & IV.

x = (3 pi)/4, (7 pi)/4; " " x = pi/3, (4 pi)/3