Решить уравнения:

а) 1 + cos 4x = cos 2x

б) 4sin^2 x - 4sin x + 1 = 0

а)

1+cos4x=cos2x

2cos ²2x=cos2x

cos2x (2cos2x-1) = 0

cos2x=0

2x=π/2+πk

x=π/4+πk/2

cos2x=1/2

2x=±π/3+2πk

x=±π/6+πk

Ответ: x=π/4+πk/2, x=±π/6+πk; k∈Z

б)

4sin²x-4sinx+1=0

(2sinx-1) ²=0

sinx=1/2

x=π/6+2πk, x=5π/6+2πk

Ответ: x=π/6+2πk, x=5π/6+2πk; k∈Z

а) 1 + cos 4x = cos 2x

2cos ²2x=cos2x

2cos ²2x-cos2x=0

cos2x (2cos2x-1) = 0

cos2x=0

2x = π/2+πk

x = π/4+πk/2, k∈z

2cos2x-1=0

cos2x=1/2

2x = - π/3+2πk U 2x=π/3+2πk

x = - π/6+πk U x=π/6+πk, k∈z

Ответ x={ π/4+πk/2; -π/6+πk; π/6+πk, k∈z}

б) 4sin^2 x - 4sin x + 1 = 0

(2sinx-1) ²=0

2sinx-1=0

sinx=1/2

x = (-1) ^k*π/6+πk, k∈z