Sin(3пx/2 -п/3)=cos(п/6-пx) РЕШИТЕ ПОЖАЛУЙСТА!!!

Решите задачу:

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