Recall #sec theta = 1/cos theta#

The secant has a vertical asymptote where the cosine of the argument is #0#.

The cosine is #0# where the argument is an odd multiple of #pi/2#

The odd multiples of #pi/2# have form #(2k+1)pi/2#, which may also be written #pi/2 + pik# for #k# an integer.

So this function has vertical asymptotes at the solutions to

#(-15pi)/8 +9x = pi/2 + pik# #" "# for #k# an integer.

#9x = (4pi)/8 -(15pi)/8 + pik# #" "# for #k# an integer.

#9x = - (11pi)/8 + pik# #" "# for #k# an integer.

#x = - (11pi)/72 + pi/8k# #" "# for #k# an integer.

**Alternatives**

We need not take #- (11pi)/72# as our base case. We can use any #x# on the list of solutions (any #x# that has the proper form).

We got #9x = - (11pi)/8 + pik# #" "# for #k# an integer.

Taking #k=2# we see that #9x = (5pi)/8# is a solution.

So we could replace the solution with

#9x = (5pi)/8 + pik# #" "# for #k# an integer.

So, the solutions are:

#x = (5pi)/72 + pi/8k# #" "# for #k# an integer.