Помогите,пожалуйста!!!!!!!! sin2x+cos(x+pi/4)=-2

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. ответ:  x=3π/4+2πk, k∈Z . * * * * * * *  sin2x+cos(x+pi/4)=-2 ;  * * * -1≤cos(x+π/4)≤1⇒ -√2≤cos(x+π/4)≤√2  ≈1,41 * ** * * * cos(x+π/4) =cosx*cosπ/4 -sinx*sinπ/4 =(√2/2)*(cosx -sinx) * * * 2sinxcosx +((√2)/2) *(cosx -sinx) = -3 +(cos²x +sin²x)  ; ((√2)/2)*(cosx - sinx) = -3  + (cos²x  - 2sinxcosx +sin²x) ; ((√2)/2)*(cosx - sinx) = -3  + (cosx  - sinx)²  2(cosx -sinx)² -√2*(cosx - sinx) -6 =0 ; * * * замена  t =cosx -sinx * * * 2t² - √2t -6=0 ; t₁=(√2 -5√2)/4 = -√2 ; t₂=(√2 +5√2)/4 =3√2 /2 . --- а) cosx -sinx= -√2 ⇔√2cos(x+π/4) = -√2⇔cos(x+π/4) = -1 ; x+π/4=π+2πk, k∈Z .  x=3π/4+2πk, k∈Z . --- б) cosx -sinx= (3√2)/2 ⇔√2cos(x+π/4) =(3√2)/2 ⇔cos(x+π/4) =3/2 >1 нет корней