## Precalculus (6th Edition) Blitzer

Please note that since $\sec (Bx-C)=\frac{1}{\cos (Bx-C)}$, the period of the secant function is the same as the period of the cosine function, that is, $\frac{2\pi }{B}$. So the period of the function $y=-\frac{3}{2} \sec \pi x$ is $\frac{2\pi }{\pi }=2$.