Solve
$\displaystyle
sec^2 2x - 1, -2\pi \leqslant x \leqslant 2\pi
$
I'm getting weird answers for this question as opposed to the actual answer.
I assume you mean $\displaystyle \sec^2(2x)-1=0$
This implies that either $\displaystyle \sec(2x)=1$ or $\displaystyle \sec(2x)=-1$
Thus, find the solutions that fall in the given interval, such that $\displaystyle x=\tfrac{1}{2}\sec^{-1}(-1)$ and $\displaystyle x=\tfrac{1}{2}\sec^{-1}(1)$
Hint...look at the values of $\displaystyle \cos^{-1}(-1)$ and $\displaystyle \cos^{-1}(1)$
--Chris