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