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

Explanation:

Step1: Recall the definition of (y = \sec(x))

We know that (\sec(x)=\frac{1}{\cos(x)}). Vertical asymptotes occur where (\cos(x) = 0).

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

The general solution of (\cos(x)=0) is (x=(2n + 1)\frac{\pi}{2}), where (n\in\mathbb{Z}) ( (n) is an integer).

• For (n = 1), (x=(2\times1 + 1)\frac{\pi}{2}=\frac{3\pi}{2})

• For (n=- 2), (x=(2\times(-2)+1)\frac{\pi}{2}=-\frac{3\pi}{2})

• For (n = 0), (x=\frac{\pi}{2})

• For (n=-1), (x=-\frac{\pi}{2})

• Check (x = - 2\pi): (\cos(-2\pi)=\cos(2\pi)=1\neq0)

• Check (x=-\frac{\pi}{6}): (\cos(-\frac{\pi}{6})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\neq0)

• Check (x=\pi): (\cos(\pi)=-1\neq0)

• Check (x = \frac{3\pi}{2}): (\cos(\frac{3\pi}{2})=0)

Answer:

(x=\frac{3\pi}{2})