General solution Find the general solution of $\sec4\theta-\sec2\theta =2$
My approach:
Converted them to cosines and then further simplified. I got $\theta = \frac{n\pi}{5}+\frac{\pi}{10}$ or $\theta = -n\pi+\frac{\pi}{2}$But my book stated the answer as $\theta = \frac{2n\pi}{5}\pm\frac{\pi}{10}$ or $\theta = 2n\pi\pm\frac{\pi}{2}$. Where am i going wrong. 
My simplification:
$$\cos2\theta - \cos4\theta = 2\cos4\theta \cdot\cos2\theta$$
$$\cos2\theta - \cos4\theta = \cos6\theta + \cos2\theta$$
$$\cos4\theta=-\cos6\theta$$
$$\cos4\theta=\cos(\pi-6\theta)$$
$$4\theta=2n\pi\pm(\pi-6\theta)$$
taking the plus and minus, I got my above-stated answer. But it doesn't quite agree with the answer given in my book. Can anyone explain my mistake?

You got the same answer, just in a different form
$$ \pm \frac{\pi}{2} + 2n\pi = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ \cdots = \frac{\pi}{2}+n\pi $$
$$ \pm \frac{\pi}{10} + \frac{2n\pi}{5} = -\frac{\pi}{10}, \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \ \cdots = \frac{\pi}{10} + \frac{n\pi}{5} $$
