Question

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

Answer 1

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

Answer 2

$\it x \in\left(\dfrac{\pi}{2},\ \pi\right)\ \ \ \ \ \ (*)\\ \\ \\ tgx+ctgx+2=0 \Leftrightarrow \dfrac{sinx}{cosx} +\dfrac{cosx}{sinx} +2=0 \Leftrightarrow sin^2+cos^2x+2sinxcosx=0\Leftrightarrow \\ \\ \\ \Leftrightarrow 1+sin2x=0 \Leftrightarrow sin2x=-1 \stackrel{(*)}{\Leftrightarrow} 2x=\dfrac{3\pi}{2}|_{:2} \Leftrightarrow x=\dfrac{3\pi}{4} \in\left(\dfrac{\pi}{2},\ \pi\right)$