Respuesta :

[tex]\cos^{2}(2x)-\sin^{2}(2x)=0[/tex]
[tex]\rightarrow \sin^{2}(2x)=\cos^{2}(2x)[/tex]
[tex]\rightarrow \frac{\sin^{2}(2x)}{\cos^{2}(2x)}=1[/tex]
[tex]\rightarrow \tan^{2}(2x)=1[/tex]
[tex]\rightarrow \tan(2x)=\pm 1[/tex]

[tex]2x[/tex] can range anywhere in [tex][0, 4\pi)[/tex]
So:
[tex]\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}[/tex]
[tex]\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}[/tex]