Решить уравнения: а)1+ cos 4x = cos 2x б)4sin^2 x - 4sin x + 1 = 0

а)
1+cos4x=cos2x
2cos²2x=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
2cos²2x=cos2x
2cos²2x-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