Question

Cum as putea rezolva acest exercitiu ???

Answer 1

$ctgx=\frac{1}{tgx} \\ \\ tgx+\frac{1}{tgx} +2=0\\ \\ tg^2x+2tgx+1=0\\ \\ (tgx+1)^2=0\\ \\ tgx=-1\\ \\ \boxed{Raspunsul~este~x\in~\{arctg(-1)+k*\pi\}~\Leftrightarrow~x\in~\{-\frac{\pi}{4} +k*\pi\},~\forall~h\in~Z.}$