Sin7x+cos квадрат 2х=sin квадрат 2х+sinx Помогите решить

sin7x+cos²2x=sin²2x+sinx (cos²2x-sin²2x)+(sin7x-sinx)=0 cos(2*2x)+2sin((7x-x)/2)*cos((7x+x)/2)=0 cos4x+2sin3x*cos4x=0 cos4x(1+2sin3x)=0 cos4x=0           или         1+2sin3x=0 4x=π/2+πn,n∈Z              2sin3x=-1 x=π/8+πn/4, n∈Z            sin3x=-1/2                                         [latex]3x=(-1)^{n+1}*\frac{\pi}{6}+\pi n, n\in Z\\x=(-1)^{n+1}*\frac{\pi}{18}+\pi n/3, n\in Z[/latex]