1+3cosx=sin2x+3sinx
Решите пожалуйста)!

Решение: 1-sin2x=cosx-sinx (cosx-sinx)²=cosx-sinx (cosx-sinx)²-(cosx-sinx )=0 (cosx-sinx)(cosx-sinx-1)=0 a) cosx-sinx=0 1-tgx=0 tgx=1 x1=π/4+πn б) cosx-sinx-1=0 cos²(x/2)-sin²(x/2)-2sin(x/2)cos(x/2)-cos²(x/2)-sin²(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=πn x2=2πn sin(x/2)-cos(x/2)=0 tg(x/2)=1 x/2=π/4+πn x3=π/2+2πn