eek, sorry that was a typo on my part

What I meant to type was

$\displaystyle \cos(2x) = 0$

$\displaystyle 2x = \pm \frac{\pi}{2} + 2\pi n$

$\displaystyle \implies x = \pm \frac{\pi}{4} + n \pi$

If we let n = 0 and sub back into the original equation we have

$\displaystyle \sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)$

$\displaystyle 1 = -1$

which is obviously wrong but I can't see where the error is

EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P