Помогите решить ур-е sinx + sin^2x/2 = cos^2x/2 и отобрать корни на промежутке[-2 пи, - пи/2]

2sinx=cos²x-sin²x

sinx=cos2x

cos2x-cos (π/2-x) = 0

-2sin (3x/2-π/4) sin (x/2+π/4) = 0

sin (3x/2-π/4) = 0 U sin (x/2+π/4) = 0

3x/2-π/4=πn U x/2+π/4=πn

3x/2=π/4+πn u x/2=-π/4+πn

x=π/6+2πn/3 U x=-π/2+2πn

x=-11π/12; -7π/12; -π/2∈[-2π; -π/2]