Cos(7x-π/8)+sin(7x-π/8)=√2. найдите х

$\cos(7x- \frac{\pi}{8} )+\sin (7x- \frac{\pi}{8})= \sqrt{2}$

Формула: $a\sin x+b\cos x = \sqrt{a^2+b^2}\sin(x+\arcsin( \frac{b}{ \sqrt{a^2+b^2} } ))$

$\sqrt{1^2+1^2} \sin(7x- \frac{\pi}{8}+\arcsin( \frac{1}{ \sqrt{1^2+1^2} } ))= \sqrt{2}\\ \\ \sqrt{2} \sin (7x- \frac{\pi}{8}+ \frac{\pi}{4})= \sqrt{2}\,\, |: \sqrt{2} \\ \\ \sin(7x+ \frac{\pi}{8} )=1\\ \\ 7x+\frac{\pi}{8}= \frac{\pi}{2}+2 \pi k,k\in Z\\ \\ 7x= \frac{3 \pi }{8} +2\pi k,k \in Z\\ \\ x= \frac{3 \pi }{56} + \frac{2 \pi k}{7} ,k \in Z$