1-sin2x=cosx-sinx,помогите,пожалуйста!

1-sin2x=cosx-sinx (cosx-sinx)²-(cosx-sinx)=0 (cosx-sinx)(cosx-sinx-1)=0 cosx-sinx=0/cosx≠0 1-tgx=0⇒tgx=1⇒x=π/4+πn,n∈Z cosx-sinx-1=0 cos²(x/2)-sin²(x/2)-2sin(x/2)cos(x/2)-sin²(x/2)-cos²(x/2)=0 2sin²(x/2)+2sin(x/2)cos(x/2)=0 2sin(x/2)(sin(x/2)-cos(x/2))=0 sin(x/2)=0⇒x/2=πk,k∈Z⇒x=2πk,k∈Z sin(x/2)-cos(x/2)=0/cos(x/2)≠0 tg(x/2)-1=0⇒tg(x/2)=1⇒x/2=π/4+πt,t∈Z⇒x=π/2+2πt,t∈Z