Сможете? Тригонометрическая

Решите задачу:

$ctg \ x-3=0\\ ctg \ x=3\\ x=arctg \frac{1}{3}+\pi n, n \in \mathbb Z \\ x= \frac{\pi}{12}+\pi n, n\in \mathbb Z$

$\sin^2{4x}-\cos^2{4x}=0\\\\ \frac{\sin^2{4x}}{\cos^2{4x}} - \frac{\cos^2{4x}}{\cos^2{4x}}=0\\\\ tg^2{4x}-1=0\\\\ tg^2{4x}=1\\\\ \frac{1-\cos8x}{\cos 8x+1}=1\\\\ 1-\cos 8x=\cos 8x+1\\\\ -\cos 8x-\cos 8x=0\\\\ -2\cos 8x=0\\\\ 8x= \frac{\pi}{2}+2\pi n, n \in \mathbb Z \\\\ x= \frac{\pi}{16}+ \frac{2\pi n}{8}, n \in \mathbb Z\\\\ x_1= \frac{\pi}{16}+ \frac{\pi n}{4}, n \in \mathbb Z\\\\ x_2=-\frac{\pi}{16}+ \frac{\pi n}{4}, n \in \mathbb Z$