Question

Find all solutions in the region [0,2pi) for the equation cos^2(2x)-sin^2(2x)=0

Answer 1

$\cos^{2}(2x)-\sin^{2}(2x)=0$
$\rightarrow \sin^{2}(2x)=\cos^{2}(2x)$
$\rightarrow \frac{\sin^{2}(2x)}{\cos^{2}(2x)}=1$
$\rightarrow \tan^{2}(2x)=1$
$\rightarrow \tan(2x)=\pm 1$

$2x$ can range anywhere in $[0, 4\pi)$
So:
$\rightarrow 2x=\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, \frac{13\pi}{4}, \frac{15\pi}{4}$
$\rightarrow x=\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8}$