## Precalculus (6th Edition)

The statement is false because $\frac{11\pi}{6}$ is not in the range of $\arcsin x$. Since $\arcsin x$ is the value of $x$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $x$, $\arcsin (-\frac{1}{2})=-\frac{\pi}{6}$.