Решите пожалуйста уровнение ни могу понять: sin10x+cos10x=√2sin6x

Решите задачу:

$sin10x+cos10x=\sqrt2\, sin6x\\\\\sqrt2\cdot ( \frac{1}{\sqrt2}\, sin10x+\frac{1}{\sqrt2} cos10x)=\sqrt2\, sin6x\\\\ \frac{1}{\sqrt2}=\frac{\sqrt2}{2}=cos \frac{\pi }{4} =sin \frac{\pi }{4}\\\\cos \frac{\pi }{4}\cdoy sin10x+sin\frac{\pi }{4}\cdot cos10x=sin6x\\\\sin(10x+\frac{\pi}{4})=sin6x\\\\sin(10x+\frac{\pi}{4})-sin6x=0\\\\2sin \frac{10x+\frac{\pi}{4}+6x}{2}\cdot cos \frac{10x+\frac{\pi}{4}-6x}{2}=0\\\\2sin(8x+\frac{\pi}{8})\cdot cos(2x+\frac{\pi}{8})=0\\\\a)\; \; sin(8x+\frac{\pi}{8})=0$

$8x+\frac{\pi}{8}=2\pi k\; ,\; \; 8x=-\frac{\pi}{8}+2\pi k\; ,\\\\x=-\frac{\pi}{64}+\frac{\pi k}{4}\; ,\; \; k\in Z\\\\b)\; \; cos(2x+\frac{\pi}{8})=0\\\\2x+\frac{\pi}{8}=2\pi n\; ,\; \; 2x=-\frac{\pi}{8}+2\pi n,\\\\x=-\frac{\pi}{16}+\pi n\; ,\; n\in Z\\\\Otvet:\; \; x=-\frac{\pi}{64}+\frac{\pi k}{4}\; ,\; \; x=-\frac{\pi }{16}+\pi n,\; \; k,n\in Z\; .$