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

1ОДЗsinx≥0⇒x∈[2πn;π+2πn,n∈z]1-cosx=sin²x(1-cosx)-(1-cosx)(1+cosx)=0(1-cosx)(1-1-cosx)=01-cosx=0cosx=1x=2πn,n∈zcosx=0x=π/2+πn,n∈z 21+сos4x=cos2x2cos²2x-cos2x=0cos2x(2cos2x-1)=0cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈zcos2x=1/2⇒2x=+-π/3+2πk⇒x=+-π/6+πk,k∈z