Помогите решить

sin^2x + cos (x-п) + 1 = 0

sin2x+cos(x+π/4)=-2 ; 

cos(x+π/4) = -3+ (cos²x + sin²x) -sin2x ;

cos(x+π/4) = -3+ (cos²x -2sinxcosx + sin²x) ;

cos(x+π/4) = -3+ (cosx  - sinx)² ;

* * * cosx -sinx = √2(cosx*1/√2 +sinx*1/√2)=√2(cosx*cosπ/4 +sinx*sinπ/4) = √2cos(x- π/4) * * *

cos(x+π/4) =-3 +(√2cos(x- π/4))² ;

2cos²(x- π/4) - cos(x+π/4) -3 =0 ;

 * * * замена t =cos(x- π/4) ; -1≤t ≤1   * * *

2t² - t -3=0 ;

t₁=(1 +5)/4 =3/2  >1 не решение исходного уравнения .

t₂=(1 -5)/4 =-1 ⇒cos(x+π/4) =-1⇔x+π/4=π+2πk, k∈Z ⇔x=3π/4+2πk, k∈Z.