Question

Solve the given equation and indicate the number of values of θ sach that 0≤θ<2π,that satisfy the iquation ;
sin(2θ)+sin(θ)=0

Answer 1

\[\sin2{\theta}+\sin{\theta}=2\sin{\theta}\cos{\theta}+\sin{\theta}=\sin{\theta}(2\cos{\theta}+1)=0\]Therefore, either \[\sin{\theta}=0\]or\[2\cos{\theta}+1=0 \rightarrow \cos{\theta}=-\frac{1}{2}\]So, if\[\sin{\theta}=0 \rightarrow \theta = 0, \pi\]and\[\cos{\theta}=-\frac{1}{2} \rightarrow{\theta}=(\pi-\frac{\pi}{3}), (\pi+\frac{\pi}{3})=\frac{2\pi}{3},\frac{4\pi}{3}\]These are the only values that solve your equation in the domain you've stipulated. Hope it helps.