# How do you solve tan(x)(tanx+1)=0?

Mar 29, 2018

$x = 0 , \frac{3 \pi}{4} , \pi , \frac{7 \pi}{4} , , 2 \pi , \frac{11 \pi}{4} , 3 \pi , \frac{15 \pi}{4} , \ldots .$

#### Explanation:

$\tan x \left(\tan x + 1\right) = 0$

$\tan x = 0$

$x = 0 , \pi , 2 \pi , 3 \pi , \ldots .$

$\tan x + 1 = 0$

$\tan x = - 1$

$x = \pi - \frac{\pi}{4} , 2 \pi - \frac{\pi}{4} , 3 \pi - \frac{\pi}{4} , 4 \pi - \frac{\pi}{4} , \ldots$

$x = \frac{3 \pi}{4} , \frac{7 \pi}{4} , \frac{11 \pi}{4} , \frac{15 \pi}{4} , \ldots .$

Combining and arranging in the increasing order

$x = 0 , \frac{3 \pi}{4} , \pi , \frac{7 \pi}{4} , , 2 \pi , \frac{11 \pi}{4} , 3 \pi , \frac{15 \pi}{4} , \ldots .$