which of the following is an asymptote of $y = \\csc(x)$?\n\\(\\bigcirc\\) $x = -\\pi$\n\\(\\bigcirc\\) $x =…

Explanation:

Step1: Recall the definition of cosecant

The cosecant function ( y = \csc(x) ) is the reciprocal of the sine function, so ( \csc(x)=\frac{1}{\sin(x)} ). A vertical asymptote occurs where the function is undefined, which is where ( \sin(x) = 0 ) (since division by zero is undefined).

Step2: Find where ( \sin(x) = 0 )

The sine function ( \sin(x) = 0 ) when ( x = n\pi ), where ( n ) is any integer (positive, negative, or zero). So the vertical asymptotes of ( y=\csc(x) ) are at ( x = n\pi ), ( n\in\mathbb{Z} ).

Step3: Check the options

• For ( x = -\pi ): Let ( n=-1 ), then ( x = (-1)\pi=-\pi ), which is of the form ( n\pi ). So ( \sin(-\pi)=0 ), so ( \csc(-\pi) ) is undefined, meaning ( x = -\pi ) is a vertical asymptote.
• For ( x = -\frac{\pi}{3} ): ( \sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\neq0 ), so ( \csc\left(-\frac{\pi}{3}\right) ) is defined, not an asymptote.
• For ( x=\frac{\pi}{4} ): ( \sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\neq0 ), so ( \csc\left(\frac{\pi}{4}\right) ) is defined, not an asymptote.
• For ( x=\frac{\pi}{2} ): ( \sin\left(\frac{\pi}{2}\right)=1\neq0 ), so ( \csc\left(\frac{\pi}{2}\right) ) is defined, not an asymptote.

Answer:

( \boldsymbol{x = -\pi} ) (corresponding to the first option: ( x = -\pi ))