Question

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

Answer 1

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

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

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