Question

which of the following is an asymptote of $y = \sec(x)$?
$\bigcirc$ $x = -2\pi$
$\bigcirc$ $x = -\frac{\pi}{6}$
$\bigcirc$ $x = \pi$
$\bigcirc$ $x = \frac{3\pi}{2}$

Explanation:

Step1: Recall sec(x) definition

$\sec(x) = \frac{1}{\cos(x)}$. Asymptotes occur where $\cos(x) = 0$.

Step2: Solve $\cos(x) = 0$

$\cos(x) = 0$ when $x = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$.

Step3: Check each option

• For $x = -2\pi$: $\cos(-2\pi) = 1

eq 0$.

• For $x = -\frac{\pi}{6}$: $\cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}

eq 0$.

• For $x = \pi$: $\cos(\pi) = -1

eq 0$.

• For $x = \frac{3\pi}{2}$: $\frac{3\pi}{2} = \frac{\pi}{2} + \pi$, so $\cos(\frac{3\pi}{2}) = 0$.

Answer:

$\boldsymbol{x = \frac{3\pi}{2}}$ (the option with $x = \frac{3\pi}{2}$)