Asymptotes for secant will be where cosine is 0
cos (6x - π/2) = 0
6x - π/2 = π/2
6x = π
x = π/6 → This is the first positive asymptote
The period of f(x) = sec x is 2π.
So the period of h(x) = 2π/6 = π/3
There are 2 asymptotes in each period of the secant function, so h(x) has an asymptote every π/6.
So the asymptotes are at \(x=\frac{\pi}{6}+\frac{k\pi}{6}\)
where k is an integer
