Am I right on this?

Question

2sec^2(2x)-8=0

2sec^2(2x)=8

divide by 2

sec^2(2x)=4

find square root

sec(2x)=2

Now, would this simply be asking me to find where secant is equal to one? Or is there no solution because secan't (pun intended) be equal to anything above 1?

Answer 1

sec^2(2x)=4

find square root
sec 2x = 2
or
sec 2x = -2