Помогите решить) 2sinxcos3x+sin4x =0

Решение 2sinxcos3x + sin4x = 0 2*(1/2)*[sin(x - 3x) + sin(x + 3x)] + sin4x = 0 - sin2x + 2sin4x = 0 2sin2x*cos2x - sin2x = 0 sin2x*(2cos2x - 1) = 0 1) sin2x = 0 2x = πk, k ∈ Z x₁ = πk/2, k ∈ Z 2)  2cos2x - 1 = 0 cos2x = 1/2 2x = (+ -)*arccos(1/2) + 2πn, n ∈ Z 2x = (+ -)*(π/3) + 2πn, n ∈ Z x₂ = (+ -)*(π/6 + πn, n ∈ Z