Question

Определить количество корней уравнения на отрезке $[ frac{- pi }{8} ; frac{ pi }{8} ]$

Само уравнение: 
$2cos (2x+frac{ pi}{3} )*sin (3x- frac{ pi}{4}) - sin (x - frac{ 7 pi}{12}) - cos ( frac{ 5 pi}{12} - 11x) = 0$

Answer 1

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