помогите решить)
2sinxcos3x+sin4x =0

Решение2sinxcos3x + sin4x = 02*(1/2)*[sin(x - 3x) + sin(x + 3x)] + sin4x = 0- sin2x + 2sin4x = 02sin2x*cos2x - sin2x = 0sin2x*(2cos2x - 1) = 01) sin2x = 02x = πk, k ∈ Zx₁ = πk/2, k ∈ Z2)  2cos2x - 1 = 0cos2x = 1/22x = (+ -)*arccos(1/2) + 2πn, n ∈ Z2x = (+ -)*(π/3) + 2πn, n ∈ Zx₂ = (+ -)*(π/6 + πn, n ∈ Z