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