On this problem, we need to figure out what the period's going to be. Regular. For cosecant is 2π. There's a coefficient in front of the X, so it's going to be 2π / 8. So the period's π force I traditionally draw in my sine graph first, so the seven tells me an amplitude of seven going down to -7. My period's π force halfway between's got to be an intercept at Pi 8's. So we know that with cosecant we can't divide by zero. So any time sign is 0, we're going to have an asymptote. So we have an asymptote at zero, another one at π eights, another one at π force. So one of the ways we can think about how to get from 1 asymptote to the next asymptote is to think about what we added zero to π eights. I added a PI8 Pi eights to π force. I added a π eights, so it's continuous everywhere except where the asymptotes occur. The asymptotes occur at north Pi 8, where N is an integer.