sin(2θ)+sin(θ)=0
use double angle formula: sin(2θ)=2sin(θ)cos(θ).
=>
2sin(θ)cos(θ)+sin(θ)=0
factor out sin(θ)
sin(θ)(2cos(θ)+1)=0
by the zero product property,
sin(θ)=0 ...........(a) or
(2cos(θ)+1)=0.....(b)
Solution to (a): θ=k(π)
solution to (b): θ=(2k+1)(π)+/-(π)/3
for k=integer
For [0,2π), this translates to:
{0, 2π/3,π,4π/3}