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

Oct 18, 2014

By rewriting a bit,

$g \left(x\right) = \sec 2 x = \frac{1}{\cos 2 x}$.

There will be vertical asymptotes when the denominator becomes 0, and $\cos 2 x$ becomes zero when

$2 x = \frac{\pi}{2} + n \pi = \frac{2 n + 1}{2} \pi$ for all integer $n$,

so, by dividing by 2,

$R i g h t a r r o w x = \frac{2 n + 1}{4} \pi$

Hence, the vertical asymptotes are

$x = \frac{2 n + 1}{4} \pi$

for all integer $n$.

I hope that this was helpful.