Find the general solution:

sinx + sin3x+ sin5x = 0

(sin x + sin 5x) + sin 3x = 0

⇒ [2 sin((x + 5x)/2) cos((x - 5x)/2)] + sin 3x = 0

[sin A + sin B = 2 sin((A + B)/2) cos((A - B)/2)]

⇒ 2sin3x cos(-2x) + sin3x = 0

⇒ 2sin3x cos 2x + sin3x = 0

⇒ sin3x (2 cos 2x + 1) = 0

⇒ sin3x = 0   or   2 cos2x + 1 = 0

Now, sin 3x = 0 ⇒ 3x = nπ, where n ∈ Ζ

i.e., x = ^nπ/₃, where n ∈ Ζ

2cos2x + 1 = 0

⇒ cos2x = -¹/₂ = cos ^π/₃ = cos(π - ^π/₃)

⇒ cos2x = cos ^2π/₃

⇒ 2x = 2nπ ± ^2π/₃, where n ∈ Ζ

⇒ x = nπ ± ^π/₃, where n ∈ Ζ