Определить количество корней уравнения на отрезке Само уравнение:

$2cos(2x+\frac{\pi}{3})*sin(3x-\frac{\pi}{4})-sin(x-\frac{7\pi}{12})-cos(\frac{5\pi}{12}-11x)=0\\\\ sin(x-\frac{7\pi}{12})=-cos(\frac{\pi}{12}-x)\\\\ 2cos(2x+\frac{\pi}{3})*sin(3x-\frac{\pi}{4})+cos(\frac{\pi}{12}-x)-cos(\frac{5\pi}{12}-11x)=0\\\\$
по формуле произведение
$2cos(2x+\frac{\pi}{3})*sin(3x-\frac{\pi}{4})=sin(5x+\frac{\pi}{12})-cos(\frac{\pi}{12}-x)\\\\ sin(5x+\frac{\pi}{12})-cos(\frac{\pi}{12}-x)+cos(\frac{\pi}{12}-x)-cos(\frac{5\pi}{12}-11x)=0\\\\ sin(5x+\frac{\pi}{12})-cos(\frac{5\pi}{12}-11x)=0\\\\ cos(\frac{5\pi}{12}-11x)=sin(11x+\frac{\pi}{12})\\\\ sin(5x+\frac{\pi}{12})-sin(11x+\frac{\pi}{12})=0\\\\ -2sin3x*cos(8x+\frac{\pi}{12})=0\\\\ x=\frac{\pi\*n}{3}\\\\ x=\frac{\pi\*n}{8}-\frac{7\pi}{96}$
Ответ 3

Определить количество корней уравнения ** отрезке Само уравнение: