How do I find the vertical asymptote of this equation

y=-csc x/3

I knwo since csc is the reciprocal of sin, the denominator (sin) cannot be 0. But I'm having trouble finding the points. is there a formula to do so ?

The cosecant function has restrictions of kπ where k is an integer.
Set kπ equal to what is in the cosecant function.
kπ = x/3
x = 3kπ
Whenever k is an integer, the equation is an asymptote. Few examples:
x = 0
x = -3π
x = 3π
x = 9π

You really don't need a formula per se, just look at the graph of -csc(x/3). Wherever -sin(x/3) equals zero is where there is going to be a vertical asymptote. And since this function is trigonometric, this means it is also periodic which means there will be vertical asymptotes every so many units backward and forward along the x-axis.

The vertical asymptotes will be where sin(x/3) = 0. Now sin(theta) = 0 for

theta = 0 + n*pi = n*pi for all integers n, so in this case since theta = x/3

the asymptotes will be where x/3 = n*pi ---> x = 3n*pi for all integers n.

So there will be asymptotes at .... -6pi, -3pi, 0, 3pi, 6pi, .... , or in degrees

....-1080, -540, 0, 540, 1080, ....

Anytime the sin graph hit's the zero line those points are considered the asymptotes of csc or basically any multiple of pi; 2pi 3pi 4 pi 5pi etc.

6pi, 12 pi, 18pi, 24pi, etc.

~David