cum ?????????????????????

tgx=t
t+1/t+2=0
t²+2t+1=0
(t+1)²=0
t=-1
tgx=-1
x=kπ+arctg(-1)=kπ+(-π/4), k∈Z
convine doar pt k=1, x=π-π/4=3π/4∈(π/2;π)
verificare
-1+(1/(-1)) +2=-1-1+2=0, adevarat
[tex] \it x \in\left(\dfrac{\pi}{2},\ \pi\right)\ \ \ \ \ \ (*)\\ \\ \\
tgx+ctgx+2=0 \Leftrightarrow \dfrac{sinx}{cosx} +\dfrac{cosx}{sinx} +2=0 \Leftrightarrow sin^2+cos^2x+2sinxcosx=0\Leftrightarrow
\\ \\ \\
\Leftrightarrow 1+sin2x=0 \Leftrightarrow sin2x=-1 \stackrel{(*)}{\Leftrightarrow} 2x=\dfrac{3\pi}{2}|_{:2} \Leftrightarrow x=\dfrac{3\pi}{4} \in\left(\dfrac{\pi}{2},\ \pi\right) [/tex]