sin2x+(1+cosx)=1
sin2x+cosx=0
2sinx*cosx+cosx=0
cosx*(2sinx+1)=0
cosx=0 ⇒ [blue]x=(π/2)+πk, k ∈ Z[/blue]
2sinx+1=0 ⇒ sinx=-1/2 ⇒ [blue]x=(-1)^(n)*(-π/6)+πn, n ∈ Z[/blue]
Отрезку [3π/2; 3π]
принадлежат корни:
[red]x_(1)=3π/2
x_(2)=(-1)^(2)*(-π/6)+2π=11π/6
x_(3)=5π/2[/red]