Решить уравнения:

а) 1 + cos 4x = cos 2x

б) 4sin^2 x - 4sin x + 1 = 0

1 + cos4x = cos2x

1 + 2cos ²2x - 1 = cos2x

2cos²2x-cos2x=0

cos2x (2cos2x-1) = 0

[cos2x=0

[cos2x=1/2

//в объединение

[2x=π/2 + πn n∈Z

[2x=±π/3 + πk k∈Z

[x=π/4 + πn/2 n∈Z

[x=±π/6 + πk/2 k∈Z

4sin^2 x - 4sin x + 1 = 0 Пусть sin x=t, |t|≤1

4t²-4t+1=0

(2t-1) ²=0

t=1/2

sinx=1/2

[x=π/6 + 2πn n∈Z

[x=5π/6 + 2πk k∈Z