sin(4x) - √3 sin(2x )= 0
2sin(2x)*cos(2x) - √3 sin(2x) = 0
sin(2x)*(2cos(2x) - √3) = 0
Имеем 2 решения:
1) sin(2x) = 0
2х = πk, k ∈ Z.
x = (π/2)k, k ∈ Z.
2) 2cos(2x) - √3 = 0
cos(2x) = √3/2
2x = (-π/6)+2πk, k ∈ Z
2x = (π/6)+2πk
x = (-π/12)+πk
x = (π/12)+πk.