Question

sin(3пx/2 -п/3)=cos(п/6-пx)
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Answer 1

$sin(frac{3pi x}{2}-frac{pi}{3})=cos(frac{pi}{6}-pi x)\ cos(frac{pi}{2}-(frac{3pi x}{2}-frac{pi}{3}))-cos(frac{pi}{6}-pi x)=0\ cos(frac{5pi}{6}-frac{3pi x}{2})-cos(frac{pi}{6}-pi x)=0\ -2*sin(frac{frac{5pi}{6}-frac{3pi x}{2}+frac{pi}{6}-pi x}{2})sin(frac{frac{5pi}{6}-frac{3pi x}{2}-frac{pi}{6}+pi x}{2})=0\ sin(frac{pi}{2}-frac{5pi x}{2})*sin(frac{pi}{3}-frac{pi x}{2})=0\ |sin(frac{pi}{2}-frac{5pi x}{2})=0\ |sin(frac{pi}{3}-frac{pi x}{2})=0$
$|sin(frac{pi}{2}-frac{5pi x}{2})=0\ |sin(frac{pi}{3}-frac{pi x}{2})=0\ \ |frac{pi}{2}-frac{5pi x}{2}=pi k\ |frac{pi}{3}-frac{pi x}{2}=pi k\ \ |frac{1}{2}-frac{5x}{2}=k\ |frac{1}{3}-frac{x}{2}=k\ \ |1-5x=2k\ |2-3x=6k\ \ |x=-frac{2}{5}k+frac{1}{5}\ |x=-2k+frac{2}{3}$