Question

Решите неравенство,пожалуйста!

$ctg^{2}$x+ctgx $\leq$ 0

Answer 1

Ответ:

$ctg^2x+ctgx\leq 0\\\\ctgx\cdot (ctgx+1)\leq 0\ \ ,\ \ \ ctgx=0\ ,\ ctgx=-1\\\\ \ \ +++[\ -1\ ]---[\ 0\ ]+++\ \ \ \ \ -1\leq ctgx\leq 0\\\\\dfrac{\pi}{2}+\pi n\leq x\leq \dfrac{3\pi}{4}+\pi n\ ,\ n\in Z\\\\Otvet:\ \ x\in \Big[\ \dfrac{\pi}{2}+\pi n\ ;\ \dfrac{3\pi}{4}+\pi n\ \Big]\ ,\ n\in Z$