Question

Решить уравнение
$\left ( \sqrt{2}-1 \right )\cos 2x+\sin 2x=1$

Answer 1

$\left ( \sqrt{2}-1 \right )\cos 2x+\sin 2x=1\Leftrightarrow \sqrt{2}\cos 2x-\left ( \cos 2x-\sin 2x \right )=1\\\sqrt{2}\cos 2x-\sqrt{2}\cos \left ( 2x+\frac{\pi}{4} \right )=1\Leftrightarrow \cos 2x-\cos \left ( 2x+\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}\\-2\sin \frac{2x+2x+\pi/4}{2}\sin \frac{2x-2x-\pi/4}{2}=1\Leftrightarrow 2\sin \frac{16x+\pi}{8}\sin \frac{\pi}{8}=\sin \frac{\pi}{4}$$2\sin \frac{16x+\pi}{8}\sin \frac{\pi}{8}=2\sin \frac{\pi}{8}\cos \frac{\pi}{8}\Leftrightarrow \sin \frac{16x+\pi}{8}=\sin \left ( \frac{\pi}{2}\pm\frac{\pi}{8} \right )\\2x+\frac{\pi}{8}=\frac{\pi}{8}+\frac{\pi}{2}\Rightarrow x=\frac{\pi}{4}+\pi k,k \in \mathbb{Z}\\2x+\frac{\pi}{8}=\frac{\pi}{2}-\frac{\pi}{8}\Rightarrow x=\frac{\pi}{8}+\pi k,k \in \mathbb{Z}$