sin2x=(1/2)sinx-(sqrt(3)/2)cosx
Вводим вспомогательный угол:
sin2x=cos(π/3)sinx-sin(π/3)cosx
sin2x=sin(x-(π/3))
sin2x-sin(x-(π/3))=0
2sin((x/2)+(π/6))*cos((3x/2)-(π/6))=0
(x/2)+(π/6)=πk, k ∈ Z
x/2=-(π/6)+πk, k ∈ Z
[b]x=-(π/3)+2πk, k ∈ Z[/b]
или
(3x/2)-(π/6)=(π/2) +πn, n ∈ Z
(3x/2)=(2π/3)+πn, n ∈ Z
[b]x=(4π/9)+(2/3)πn, n ∈ Z[/b]