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