We have
sin θ + sin 3θ + sin 5θ = 0
or, (sin θ + sin 5θ) + sin 3θ = 0
or, 2 sin 3θ cos 2θ + sin 3θ = 0
or, sin 3θ (2 cos 2θ + 1) = 0
or, sin 3θ = 0 or cos 2θ = –1/2
When sin 3θ = 0, then 3θ = nπ or θ =nπ/3
When cos 2θ = –1/2 = cos2π/3,
then 2θ = 2nπ ± 2π/3 or, θ = nπ ± π/3
which gives
All these values of θ are contained in θ = nπ/3, n ∈ Z. Hence, the required solution set is given by {θ : θ = nπ/3, n ∈ Z}
