# For what values of x, if any, does f(x) = sec((-11pi)/6-7x)  have vertical asymptotes?

Dec 16, 2017

$\implies x = \frac{- 2 \pi n}{7} \pm - \frac{\pi}{14} - \frac{11 \pi}{42}$

$n \in \mathbb{Z}$

#### Explanation:

We need to consider the definition of:

$\sec x = \frac{1}{\cos} x$

Hence:

$\sec \left(\frac{- 11 \pi}{6} - 7 x\right) = \frac{1}{\cos \left(\frac{- 11 \pi}{6} - 7 x\right)}$

Hence there are verticle asymtptotes where the denominator $= 0$

$\implies \cos \left(\frac{- 11 \pi}{6} - 7 x\right) = 0$

$\implies \cos \left(\frac{- 11 \pi}{6} - 7 x\right) = \cos \left(\frac{\pi}{2}\right)$

Using the general solution for $\cos x$:

If $\cos x = \cos \phi$
$\implies x = 2 \pi n \pm \phi$

$\implies \frac{- 11 \pi}{6} - 7 x = 2 \pi n \pm \frac{\pi}{2}$

$\implies - 7 x = 2 \pi n \pm \frac{\pi}{2} + \frac{11 \pi}{6}$

$\implies x = \frac{- 2 \pi n}{7} \pm - \frac{\pi}{14} - \frac{11 \pi}{42}$