Решить уравнение:

1-cos4x=2sin2x

1- cos4x = 2sin2x2sin²2x - 2sin2x = 0sin2x(sin2x - 1) = 01) sin 2x = 0   → 2x = πk → x = πk/2   (k ∈ Z)2) (sin2x - 1) = 0 → sin2x = 1 → 2x = π/2 + 2πk  → x = π/4 + πk  (k ∈ Z)Ответ: x1 = πk/2   (k ∈ Z);   x2 = π/4 + πk  (k ∈ Z)