Să se rezolve ecuatia :
sinx+cosx=
[tex] \sqrt{2} [/tex]
tgx=?

[tex]sin(x)+cos(x)=\sqrt2\ \Bigr|^2[/tex]
[tex]sin^2(x)+2sin(x)cos(x)+cos^2(x)=2[/tex]
[tex]1+sin(2x)=2[/tex]
[tex]sin(2x)=1[/tex]
[tex]2x=2n\pi+\cfrac{\pi}{2},\ n\in\mathbb{Z}[/tex]
[tex]x=n\pi+\cfrac{\pi}{4},\ n\in\mathbb{Z}[/tex]
Dar [tex]x\in(0,\ \cfrac{\pi}{2})[/tex]
Deci singura solutie este atunci cand [tex]n=0[/tex] si anume
[tex]{x=\cfrac{\pi}{4}[/tex]
Deci
[tex]\boxed{tg(x)=tg\left(\cfrac{\pi}{4}\right)=1}[/tex]