## Precalculus (6th Edition)

$s=\dfrac{11\pi}{6}$
Use the inverse cosine function of a calculator in radian mode to obtain: $s =\cos^{-1}{(\frac{\sqrt3}{2})}=\frac{\pi}{6}$ The angle must be within the interval $[\frac{3\pi}{2}, 2\pi]$ Note $\cos{(\frac{\pi}{6})}=\cos{(-\frac{\pi}{6})}$. Since $\cos{\theta} = \cos{(\theta+2\pi)}$, then $\cos{(-\frac{\pi}{6})}=\cos{(-\frac{\pi}{6}+2\pi)}=\cos{(\frac{11\pi}{6})}$ Thus, $s=\dfrac{11\pi}{6}$