Where are the asymptotes of f (x) = tan (4x - π) from x = 0 to x = pi over 2?

Asymptotes of tan (x) are at (k+1/2) π. (k=any integer)

So the asymptotes of f (x) are

f (x) = tan (4x-π ) = tan ((k+1/2) π )

=>

4x-π = (k+1/2) π

= (k+1/2) π +π

= (k+1/2) π

=>

x = (k+1/2) π /4

= (k/4+1/8) π

For

k=-1, x=-π /8 [outside of (0,π /2) ]

k=0, x=π /8

k=1, x=3π /8

k=2, x=5π /8 [outside of (0,π /2) ]

So the answer is x={π /8, 3π /8}