Решите уравнение. 4sin(в 4 степени)2x+3cos4x-1=0

Решите задачу:

$4\sin^4 2x + 3\cos 4x - 1 = 0\\\\ \left[ \ \cos 2x = \cos^2 x - \sin^2 x \ \right]\\\\ 4\sin^4 2x + 3(\cos^2 2x - \sin^2 2x) - 1 = 0\\\\ 4\sin^4 2x + 3\cos^2 2x - 4\sin^2 2x + \sin^2 2x - 1 = 0\\\\ \left[ \ \sin^2x - 1 = -\cos^2x \ \right]\\\\ 4\sin^2 2x(\sin^2 2x - 1) + 3\cos^2 2x - \cos^2 2x = 0\\\\ - 4\sin^2 2x \cos^2 2x + 2 \cos^2 2x = 0\\\\ 2\cos^2x (1 - 2\sin^2 2x) = 0$

$\left[ \ 1 - 2\sin^2 x = \cos 2x \ \right]\\\\ 2\cos^2 2x \cos 4x = 0\\\\ 1. \ \cos 2x = 0\\\\ 2x = \pi n + \frac{\pi}{2}, \ n \in \mathbb{Z}\\\\ \boxed{x = \frac{\pi}{2}n + \frac{\pi}{4}, \ n \in \mathbb{Z}}\\\\ 2. \ cos 4x = 0\\\\ \boxed{x = \frac{\pi}{4}n + \frac{\pi}{8}, \ n \in \mathbb{Z}}$