## anonymous 5 years ago 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

$\sin2{\theta}+\sin{\theta}=2\sin{\theta}\cos{\theta}+\sin{\theta}=\sin{\theta}(2\cos{\theta}+1)=0$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.