## Precalculus (6th Edition) Blitzer

The sine functions in the two equations have different periodicity. For $sin(x)=\frac{1}{2}$, the period is $2\pi$ and there are two solutions in $[0,2\pi$. While for $sin(x)=\frac{1}{2}$, because the period is $\pi$, the number of solutions will double to four.