Решите уровнение 1+cos4x=cos2x

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