помогите решить уравнение 2cos^2 x + sin2x = sin(x -3п/2) - cos(п/2 +x)

sin(3π/2-a)=-cosa ⇒sin(x-3π/2)=-sin(3π/2-x)=cosxcos(π/2+x)=-sinx    sin2x=2sinx*cosx2cos²x+2sinxcosx=cosx+sinx2cosx*(cosx+sinx)-(cosx+sinx)=0(cosx+sinx)*(2cosx-1)=0cosx+sinx=0/cosx1+tgx=0tgx=-1x=-π/4+πn,n∈z2cosx-1=02cosx=1cosx=1/2x=+-π/3+2πk,k∈z