Question 592892
We will first find a general solution for all the asymptotes. Then we will use that to find the specific ones the problem asks you to find.<br>
The (vertical) asymptotes will occur for x values that make {{{sec(x - pi/2)}}} undefined. Since sec is the reciprocal of cos, sec is undefined when cos is zero. So we are interested in the solution to:
{{{cos(x-pi/2) = 0}}}
You might be able to figure this out in your head. If not, we can make it easier to solve by using the cos(A-B) formula:
cos(A-B) = cos(A)*cos(B) + sin(A)sin(B)
With A = x and B = {{{pi/2}}},
{{{cos(x-pi/2) = 0}}}
becomes
{{{cos(x)*cos(pi/2) + sin(x)*sin(pi/2) = 0}}}
The cos and sin of {{{pi/2}}} are known. Substituting these values in we get:
cos(x)*0 + sin(x)*1 = 0
which simplifies to:
sin(x) = 0<br>
This is easily solved. Since sin(x) = 0 at 0 and at {{{pi}}}, the general solution is
x = {{{0 + 2pi*n}}}   (where "n" is any integer)
or
x = {{{pi + 2pi*n}}}   (where "n" is any integer)<br>
From the general solution above, we can now find the desired specific solutions. Just play around with different integer values for "n" until you find the two smallest non-negative asymptotes and the first negative asymptote. (Reread the start of this solution to remind yourself why these solutions to sin(x)=0 turn out to be asymptotes for your original equation.)<br>
(You'll find that using n=0 in the first equation and n=0 in the second equation will give you the two smallest non-negative asymptotes and using n = -1 in the second equation will give you the first negative asymptote.)