Sin4x = cos6x x - ? - Все ответы

Решите задачу:

$\sin 4x = \sin(\frac{\pi}{2}+6x) \\ \\ \sin 4x - \sin(\frac{\pi}{2}+6x)=0 \\ \\ 2 \sin{(\frac{4x-\frac{\pi}{2} - 6x }{2})}\cdot \cos {(\frac{4x+\frac{\pi}{2} + 6x }{2})}=0\\\\2 \sin{(\frac{-2x-\frac{\pi}{2} }{2})}\cdot \cos {(\frac{10x + \frac{\pi}{2}}{2})}; \ \ \ \ -2 \sin{(\frac{2x+\frac{\pi}{2} }{2})}\cdot \cos {(\frac{10x + \frac{\pi}{2}}{2})}=0; \\\\ \sin{(\frac{2x+\frac{\pi}{2} }{2})}\cdot \cos {(\frac{10x + \frac{\pi}{2}}{2})}=0$

$\sin{(\frac{2x+\frac{\pi}{2} }{2})}=0 \\ \\ \frac{2x+\frac{\pi}{2} }{2}=\pi n, \ n \in Z \\ \\ 2x+\frac{\pi}{2}=2\pi n, \ n \in Z \\ \\ 2x=-\frac{\pi}{2} + 2\pi n, \ n \in Z \\ \\ \boxed{x=-\frac{\pi}{4} + \pi n, \ n \in Z }$

$\cos {(\frac{10x + \frac{\pi}{2}}{2})}=0 \\ \\ \frac{10x + \frac{\pi}{2}}{2}=\frac{\pi}{2}+ \pi n, \ n \in Z \\ \\ 10x + \frac{\pi}{2}=\pi+ 2\pi n, \ n \in Z \\ \\ 10x =\pi- \frac{\pi}{2}+ 2\pi n, \ n \in Z \\ \\ 10x =\frac{\pi}{2}+ 2\pi n, \ n \in Z \\ \\ \boxed{x =\frac{\pi}{20}+ \frac{\pi n}{5}, \ n \in Z}$