Решить 2cos^2(3pi/2+x)=sin2x -9pi/2 ; -3pi]

2cos²(3π/2+x)=sin2x
2sin²x-2sinxcosx=0
2sinx(sinx-cosx)=0
sinx=0      sinx-cosx=0
x=πn,n∈Z; tgx-1=0
                 tgx=1
                 x=π/4+πn,n∈Z