Trigonometry help plz!

Find the equations for all vertical asymptotes for the function.

y = csc (4x + pi)

Answer: x = kpi/4

I already the answer but how did you get this? Could you plz explain the steps?

A vertical line on a graph has an equation x = C where C is some constant.

Vertical asymptotes are those lines where y would be undefined and thus cannot exist.

Since the csc(v) = 1/sin(v), the vertical asymptote is anywhere where sin(v) = 0. The sin(v) is zero wherever v = 0, pi, 2pi, 3pi or more generally, m(pi) where m is an integer.

4x + pi = mpi
4x = (m-1)pi
x = (m-1)pi/4

But this is the same form of the above answer. If m-1 is an integer, then k = m-1 is also an integer.

so it is okay to set x = kpi/4

Jim

Your function y = csc (4x + pi) is the same as 1/[sin (4x+π)]. Your vertical asymptotes are those vertical lines that are at the limit as you let the denominator ==> zero.

The denom => zero when the sine argument is 0+nπ (where n is any integer.

Setting the argument to this solution: 4x+π = 0+nπ which if you solve for x gives

x = [0-nπ-π]/4 But since n is ANY integer, the bracket quantity is also generated by the simpler
expression nπ/4 (which ISN'T MORE CORRECT, but is simpler!).