Sin3x + sin5x = sin4x
2[sin((3x + 5x)/2)·cos(3x - 5x)/2)] = sin4x
2sin4xcosx - sin4x = 0
sin4x(2cosx - 1) = 0
sin4x = 0
4x = πn, n ∈ Z
x = πn/4, n ∈ Z
2cosx - 1 = 0
2cosx = 1
cosx = 1/2
x = ±π/3 + 2πk, k ∈ Z
Ответ: x = πn/4, n ∈ Z; ±π/3 + 2πk, k ∈ Z.