Реши уравнение: cos4x-sin 3x = 0

Решим уравнения: 1) $\cos 4x - \sin 3x = 0$; $\cos 4x = \sin 3x$; $\cos 4x = \cos (\frac{\pi}{2} - 3x)$; $4x = \pm (\frac{\pi}{2} - 3x) + 2\pi k, k \in \mathbb{Z}$; a) $4x = \frac{\pi}{2} - 3x + 2\pi k$; $7x = \frac{\pi}{2} + 2\pi k$; $x = \frac{\pi}{14} + \frac{2\pi k}{7}, k \in \mathbb{Z}$; b) $4x = -\frac{\pi}{2} + 3x + 2\pi k$; $x = -\frac{\pi}{2} + 2\pi k, k \in \mathbb{Z}$. 2) $\sin (\frac{\pi}{4} - x) - \cos (\frac{\pi}{4} - x) = -1$; $\sin (\frac{\pi}{4} - x) - \sin (\frac{\pi}{2} - (\frac{\pi}{4} - x)) = -1$; $\sin (\frac{\pi}{4} - x) - \sin (\frac{\pi}{4} + x) = -1$; $2 \cos \frac{(\frac{\pi}{4} - x + \frac{\pi}{4} + x)}{2} \sin \frac{(\frac{\pi}{4} - x - \frac{\pi}{4} - x)}{2} = -1$; $2 \cos \frac{\pi}{4} \sin (-x) = -1$; $2 \cdot \frac{\sqrt{2}}{2} \cdot (-\sin x) = -1$; $\sin x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$; $x = \frac{\pi}{4} + 2\pi k, k \in \mathbb{Z}$; $x = \frac{3\pi}{4} + 2\pi k, k \in \mathbb{Z}$. 3) $\sin 2x + \sin 8x = \sqrt{2} \cos 3x$; $2 \sin \frac{2x + 8x}{2} \cos \frac{2x - 8x}{2} = \sqrt{2} \cos 3x$; $2 \sin 5x \cos (-3x) = \sqrt{2} \cos 3x$; $2 \sin 5x \cos 3x = \sqrt{2} \cos 3x$; $2 \sin 5x \cos 3x - \sqrt{2} \cos 3x = 0$; $\cos 3x (2 \sin 5x - \sqrt{2}) = 0$; a) $\cos 3x = 0$; $3x = \frac{\pi}{2} + \pi k$; $x = \frac{\pi}{6} + \frac{\pi k}{3}, k \in \mathbb{Z}$; b) $2 \sin 5x - \sqrt{2} = 0$; $\sin 5x = \frac{\sqrt{2}}{2}$; $5x = \frac{\pi}{4} + 2\pi k$; $x = \frac{\pi}{20} + \frac{2\pi k}{5}, k \in \mathbb{Z}$; $5x = \frac{3\pi}{4} + 2\pi k$; $x = \frac{3\pi}{20} + \frac{2\pi k}{5}, k \in \mathbb{Z}$. 4) $\sin x + \sin 7x = \sin 5x + \sin 3x$; $\sin x + \sin 7x - \sin 5x - \sin 3x = 0$; $(\sin x - \sin 3x) + (\sin 7x - \sin 5x) = 0$; $2 \cos \frac{x + 3x}{2} \sin \frac{x - 3x}{2} + 2 \cos \frac{7x + 5x}{2} \sin \frac{7x - 5x}{2} = 0$; $2 \cos 2x \sin (-x) + 2 \cos 6x \sin x = 0$; $-2 \cos 2x \sin x + 2 \cos 6x \sin x = 0$; $2 \sin x (\cos 6x - \cos 2x) = 0$; $2 \sin x (-2 \sin \frac{6x + 2x}{2} \sin \frac{6x - 2x}{2}) = 0$; $-4 \sin x \sin 4x \sin 2x = 0$; a) $\sin x = 0$; $x = \pi k, k \in \mathbb{Z}$; b) $\sin 4x = 0$; $4x = \pi k$; $x = \frac{\pi k}{4}, k \in \mathbb{Z}$; c) $\sin 2x = 0$; $2x = \pi k$; $x = \frac{\pi k}{2}, k \in \mathbb{Z}$.