# Thread: How do you do this? - Trigonometric Equation

1. ## How do you do this? - Trigonometric Equation

Solve

$
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.

2. Originally Posted by cdx
Solve

$
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 $\sec^2(2x)-1=0$

This implies that either $\sec(2x)=1$ or $\sec(2x)=-1$

Thus, find the solutions that fall in the given interval, such that $x=\tfrac{1}{2}\sec^{-1}(-1)$ and $x=\tfrac{1}{2}\sec^{-1}(1)$

Hint...look at the values of $\cos^{-1}(-1)$ and $\cos^{-1}(1)$

--Chris