1)Cos3x=sin7x 2)(sin 11pi/18 - sin pi/18)/(cos 11pi/18 - cos pi/18)

Решите задачу:

$cos3x=sin7x\\sin(\frac{\pi}{2}-3x)-sin7x=0\\2sin(\frac{\pi}{4}-5x)cos(\frac{\pi}{4}+2x)=0\\\\sin(\frac{\pi}{4}-5x)=0\\\frac{\pi}{4}-5x=\pi n\\-5x=-\frac{\pi}{4}+\pi n\\x=\frac{\pi}{20}-\frac{\pi n}{5}, \; n\in Z;\\\\cos(\frac{\pi}{4}+2x)=0\\\frac{\pi}{4}+2x=\frac{\pi}{2}+\pi k \\2x=\frac{\pi}{4}+\pi k\\x=\frac{\pi}{8}+\frac{\pi k}{2}, \; k\in Z.$

$\frac{sin\frac{11\pi}{18}-sin\frac{\pi}{18}}{cos\frac{11\pi}{18}-cos\frac{\pi}{18}} =\frac{2sin\frac{5\pi}{18}cos\frac{\pi}{3}}{-2sin\frac{\pi}{3}sin\frac{5\pi}{18}}=-\frac{cos\frac{\pi}{3}}{sin\frac{\pi}{3}}=-ctg\frac{\pi}{3}=-\frac{\sqrt 3}{3}.$