РЕШИТЕ УРАВНЕНИЯ 1)корень из (1-cosx)=sinx 2)1+cos4x =cos2x

1 ОДЗ sinx≥0⇒x∈[2πn;π+2πn,n∈z] 1-cosx=sin²x (1-cosx)-(1-cosx)(1+cosx)=0 (1-cosx)(1-1-cosx)=0 1-cosx=0 cosx=1 x=2πn,n∈z cosx=0 x=π/2+πn,n∈z 2 1+сos4x=cos2x 2cos²2x-cos2x=0 cos2x(2cos2x-1)=0 cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈z cos2x=1/2⇒2x=+-π/3+2πk⇒x=+-π/6+πk,k∈z