What are the asymptotes of #g(x)=sec 2x#?

1 Answer
Oct 18, 2014

By rewriting a bit,

#g(x)=sec2x=1/{cos2x}#.

There will be vertical asymptotes when the denominator becomes 0, and #cos2x# becomes zero when

#2x=pi/2+npi={2n+1}/2pi# for all integer #n#,

so, by dividing by 2,

#Rightarrow x={2n+1}/4pi#

Hence, the vertical asymptotes are

#x={2n+1}/4pi#

for all integer #n#.


I hope that this was helpful.