How do you find the x intercepts for #y=sec^2x-1#?

1 Answer
Aug 1, 2016

#pi+pin=x#

Explanation:

Note that #sec^2x-1=tan^2x#. Hence, we can replace the original equation for a simpler one. Albeit working with #sec# is also easy in this case, it's a good habit to work with fewer terms. (There are occasions in which this is not true, but that will with experience.)

Also, note that the only way for #tan^2x=0# is when #tanx=0#.
So, when is it zero? If you remember the unit circle, you can see that #tanx=o/a#, where #o# is the opposite side in the unit circle, and #a# the adjacent.

Hence, you want #o=0#. And that happens when #pi+pin=x#, where #pi# is any integer.
Here's a graphical representation:
graph{tanx [-10, 10, -5, 5]}