Răspuns:
3.
3sinx+√3cosx=0║:√3
√3sinx+cosx=0
√3=tgπ/3 inlocuiesti in ecuatie
tgπ/3*sinx+cosx=0
(sinπ/3/cosπ/3)*sinx+cosx=0
sinπ/3*sinx+cosx*cosπ/3=0<=>
cosx*cosπ/3+sinπ/3*sinx=0
se aplica formula cosαcosβ+sin αsinβ=cos(α-β)
cos(x-π/3)=0=>
x-π/3=π/2
x=π/2+π/3
x=5π/6
x=5π/6+2kπ
Explicație pas cu pas:
