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)=cosx
cos(π/2+x)=-sinx   
sin2x=2sinx*cosx
2cos²x+2sinxcosx=cosx+sinx
2cosx*(cosx+sinx)-(cosx+sinx)=0
(cosx+sinx)*(2cosx-1)=0
cosx+sinx=0/cosx
1+tgx=0
tgx=-1
x=-π/4+πn,n∈z
2cosx-1=0
2cosx=1
cosx=1/2
x=+-π/3+2πk,k∈z