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