Question

Alguém sebe resolver?

tgx(tgx + 1)=0

Answer 1

$\displaystyle \tan x=0 \vee \tan x = -1\\ \\ x\in \{k\pi:k\in\mathbb Z\} \vee x\in \left\{\frac{3\pi}{4}\pm 2k\pi ; \frac{7\pi}{4}\pm2k\pi:k\in\mathbb Z\right\}\\ \\ \boxed{x \in \left\{k\pi\,;\,\frac{3\pi}{4}\pm 2k\pi \,;\, \frac{7\pi}{4}\pm2k\pi:k\in\mathbb Z\right\}}$