Как решить тригонометрическое уравнение (помогите, плиииз...)sinx = cos7x ?

Решите задачу:

$cos \alpha =sin(\frac{\pi}{2}- \alpha )$

$sinx=cos7x\\\\sinx-sin(\frac{\pi}{2}-7x)=0\\\\2sin\frac{x-\frac{\pi}{2}+7x}{2}\cdot cos\frac{x+\frac{\pi}{2}-7x}{2}=0\\\\sin(4x-\frac{\pi}{4})\cdot cos(\frac{\pi}{4}-3x)=0\\\\1.\; sin(4x-\frac{\pi}{4})=0,\\\\4x-\frac{\pi}{4}=\pi n,\; \; 4x=\frac{\pi}{4}+\pi n\; ,\; \; x=\frac{\pi}{16}+\frac{\pi n}{4},n\in Z\\\\2.\; cos(\frac{\pi}{4}-3x)=0,\\\\\frac{\pi}{4}-3x=\frac{\pi}{2}+\pi k,\; \; 3x=-\frac{\pi}{4}-\pi k,\; \; x=-\frac{\pi}{12}-\frac{\pi k}{3},k\in Z\\\\Mozno\; tak:\; x=-\frac{\pi}{12}+\frac{\pi k}{3}$