## Precalculus (6th Edition) Blitzer

The statement is false. The correct statement is: the equations $\sin 2x=1$ and $\sin 2x=\frac{1}{2}$ do not have the same number of solutions on the interval $[0,2\pi )$.
Solution of $\sin 2x=1$ will be $\frac{\pi }{4}$ and $\frac{5\pi }{4}$ So, the number of solutions will be $2$. Solutions of $\sin 2x=\frac{1}{2}$ will be $\frac{\pi }{12}$ , $\frac{5\pi }{12}$ , $\frac{13\pi }{12}$ , $\frac{17\pi }{12}$ Therefore, the number of solutions will be $4$. So, the number of solutions on the interval $[0,2\pi )$ are not the same.