Question

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

Answer 1

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


$frac{sinfrac{11pi}{18}-sinfrac{pi}{18}}{cosfrac{11pi}{18}-cosfrac{pi}{18}} =frac{2sinfrac{5pi}{18}cosfrac{pi}{3}}{-2sinfrac{pi}{3}sinfrac{5pi}{18}}=-frac{cosfrac{pi}{3}}{sinfrac{pi}{3}}=-ctgfrac{pi}{3}=-frac{sqrt 3}{3}.$