Пожалуйста решить cos^2x+cos^2y-sin^2(x+y)=2cosx

cos²x +cos²y -sin²(x+y) = 2cosx  ; (1+cos2x)/2 +(1+cos2y)/2 -(1-cos2(x+y))/2 = 2cosx  ; 1+cos2x +1+cos2y -1+cos2(x+y) = 4cosx  ; (1+cos2(x+y) ) +(cos2x +cos2y )= 4cosx  ; 2cos²(x+y) +2cos(x+y)cos(x-y) = 4cosx  ; 2cos(x+y)( cos(x+y)+cos(x-y)) = 4cosx ; 2cos(x+y)*2 cosx*cosy = 4cosx ; 4cosx (cos(x+y)cosy -1) =0 ; а) cosx =0 ; x =π/2 +πk , k∈Z . б) cos(x+y)cosy -1 =0 ⇔ cos(x+y)cosy=1 . б₁)  {cos(x+y) = -1 ; cosy= -1. { x+y =π+2πk ; y = π+2πn ⇒{x=2π(k -n) ; y = π+2πn . б₂)  {cos(x+y) =1 ; cosy= 1 ; {x+y =2πk ; y = 2πn ⇒{x=2π(k -n) ; y = 2πn .