Question

1.cos(3x-п/6)+cos(x+п/4)=0. 2.cos(2x+п/4)+cosx/2=0

Answer 1

$1. \; cos(3x-\frac{\pi}{6})+cos(x+\frac{\pi}{4})=0\\2cos\frac{3x-\frac{\pi}{6}+x+\frac{\pi}{4}}{2}cos\frac{3x-\frac{\pi}{6}-x-\frac{\pi}{4}}{2}=0\\2cos(2x+\frac{\pi}{24})cos(x-\frac{5\pi}{24})=0\\\\cos(2x+\frac{\pi}{24})=0\\2x+\frac{\pi}{24}=\frac{\pi}{2}+\pi n\\2x=\frac{11\pi}{24}+\pi n\\x=\frac{11\pi}{48}+\frac{\pi n}{2}, \; n\in Z;\\\\cos(x-\frac{5\pi}{24})=0\\x-\frac{5\pi}{24}=\frac{\pi}{2}+\pi n\\x=\frac{17\pi}{24}+\pi n, \; n\in Z.$

$2. \; cos(2x+\frac{\pi}{4})+cos\frac{x}{2}=0\\2cos\frac{2x+\frac{\pi}{4}+\frac{x}{2}}{2}cos\frac{2x+\frac{\pi}{4}-\frac{x}{2}}{2}=0\\2cos(\frac{5x}{4}+\frac{\pi}{8})cos(\frac{3x}{4}+\frac{\pi}{8})=0\\\\cos(\frac{5x}{4}+\frac{\pi}{8})=0\\\frac{5x}{4}+\frac{\pi}{8}=\frac{\pi}{2}+\pi n\\\frac{5x}{4}=\frac{3\pi}{8}+\pi n\\x=\frac{3\pi}{10}+\frac{4\pi n}{5}, \; n\in Z;$

$cos(\frac{3x}{4}+\frac{\pi}{8})=0\\\frac{3x}{4}+\frac{\pi}{8}=\frac{\pi}{2}+\pi n\\\frac{3x}{4}=\frac{3\pi}{8}+\pi n\\x=\frac{\pi}{2}+ \frac{4\pi n}{3}, \; n\in Z.$