1+cos4x=cos2x
cos4x=cos(2*(2x))=2cos²2x-1
1+(2cos²2x-1)=cos2x
2cos²2x-2cos2x=0
cos2x*(2cos2x-1)=0
cos2x=0 или 2cos2x-1=0
1. cos2x=0, 2x=π/2+πn, n∈Z. x=π/4+πn/2, n∈Z
2. 2cos2x-1=0, cos2x=1/2. 2x=+-arccos(1/2)+2πn, n∈Z. 2x=+-π/3+2πn, n∈Z. x=+-π/6+πn, n∈Z
ответ: x₁=π/4+πn/2, n∈Z
x₂=+-π/6+πn, n∈Z