## Algebra 2 (1st Edition)

$\dfrac{2\sqrt 3}{3}$
When $P$ is the period of $f(x)$ then we have $f(x+p)=f(x)$ Here, we have $\theta= \dfrac{11 \pi}{6}$ lies in the fourth quadrant thus, the angle is in the interval $[0, 2 \pi]$. But $\theta'=2 \pi-\dfrac{11 \pi}{6}=\dfrac{\pi}{6}$ and $\sec \theta'=\sec \dfrac{\pi}{6}=\dfrac{1}{\cos \dfrac{\pi}{6}}$ $\dfrac{1}{\cos \dfrac{\pi}{6}}=\dfrac{1}{\sqrt 3/2}$ or, $=\dfrac{2}{\sqrt 3}$ or, $=\dfrac{2\sqrt 3}{3}$