Question

1.sin11п/18-sinп/18 /cos 11п/18-cos п/18 2.sin 5x-sin 6x=0 3.cos5x=cos 7x

Answer 1

$1)\; \; \frac{sin \frac{11\pi}{18} -sin\frac{\pi}{18}}{cos\frac{11\pi}{18} -cos\frac{\pi }{18}}= \frac{2sin\frac{5\pi}{18}\cdot cos\frac{\pi}{3}}{-2sin\frac{5\pi}{18}\cdot sin\frac{\pi}{3} } =-ctg\frac{\pi}{3}=-\frac{\sqrt3}{3}\\\\2)\; \; sin5x-sin6x=0\\\\2sin\frac{5x-6x}{2}\cdot cos\frac{5x+6x}{2}=0\\\\sin(-\frac{x}{2})\cdot cos\frac{11x}{2}=0\\\\-sin\frac{x}{2}\cdot cos\frac{11x}{2}=0\\\\a)\; \; sin\frac{x}{2}=0\\\\ \frac{x}{2} =\pi n,\; n\in Z\\\\x=2\pi n,; n\in Z\\\\b)\; \; cos\frac{11x}{2}=0$

$\frac{11x}{2}= \frac{\pi}{2}+\pi k,\; k\in Z\\\\x=\frac{\pi}{11}+\frac{2\pi k}{11},\; k\in Z\\\\3)\; \; cos5x=cos7x\\\\cos5x-cos7x=0\\\\-2sin(-x)\cdot sin6x=0\\\\a)\; \; sin(-x)=-sinx=0\; ,\; \; \; x=\[i n,\; n\in Z\\\\b)\; \; sin6x=0\; ,\; \; 6x=\pi k,\; k\in Z\\\\x=\frac{\pi k}{6}\; ,\; k\in Z$