Question

Помогите решить 1)sin(2x+п/7)=^3/2
2)sin(x-3)=1
3)sin5x=0
4)sin(4x-п/3)=-1

Answer 1

1)2x+π/7=(-1)^n*π/3+πn
2x=(-1)^n*π/3-π/7+πn
x=(-1)^n*π/6-π/14+πn/2
2)x-3=π/2+2πn
x=π/2+3+πn
3)5x=πn
x=πn/5
4)4x-π/3=-π/2+2πn
4x=-π/2+π/3+2πn=-π/6+2πn
x=-π/24+πn/2