cos2x=sinx-cosx тригонометрическое уравнение

cos²x-sin²x-(sinx-cosx)=0(cosx-sinx)(cosx+sinx)+(cosx-sinx)=0(cosx-sinx)(cosx+sinx+1)=0cosx-sinx=0/cosx1-tgx=0tgx=1x=π/4+πn,n∈zcosx+sinx+1=02cos²(x/2)+2sin(x/2)cos(x/2)=02cos(x/2)*(cos(x/2)+sin(x/2)=0cos(x/2)=0x/2=π/2+πkx=π+2πk,k∈zcos(x/2)+sin(x/2)=0/cos(x/2)1+tg(x/2)=0tg(x/2)=-1x/2=-π/4+πmx=-π/2+2πm,m∈z