How would you write equations for three vertical asymptotes of the function y= Sec X

x = n*pi where n is some integer, just pick three integers.  

Step-by-step explanation:

Vertical asymptotes are vertical, so to write the equation of a vertical line, much like a horizontal is y = a, a vertical line is x=b.  So you just need to find where the vertical asymptotes are.

sec(x) is 1/cos(x).  the VAs occure when cos(x) is 0?  well, when does it equal 0?  Basically at 0 then every pi along the x axis positive or negative afterwards.   so basically n*pi where n is some integer, so that's negatves, 0 and positives.  just choose three integers.

EDIT

I was wrong.  cos(x) is 0  when x = pi*n + pi/2 Or in other words n*pi/2 where n is an odd integer.  I mistakenly explained when sin(x) = 0.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

i think it is 70

step-by-step explanation: