Question

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Answer 1

$1)\; \; sin(\frac{\pi }{6}-3x)-\frac{1}{2}=0\\\\sin(\frac{\pi}{6}-3x)=\frac{1}{2}\\\\\frac{\pi }{6}-3x=(-1)^{n}\cdot \frac{\pi }{6}+\pi n\; ,\; n\in Z\\\\3x=\frac{\pi}{6}-(-1)^{n}\cdot \frac{\pi }{6}-\pi n\; ,\; n\in Z\\\\x=\frac{\pi }{18}-(-1)^{n}\cdot \frac{\pi }{18}-\frac{\pi n}{3}\; ,\; n\in Z\\\\2)\; \; 2cos(\frac{1}{3}x-10\pi )=\sqrt3\\\\cos(\frac{x}{3}-10\pi )=\frac{\sqrt3}{2}\\\\\frac{x}{3}-10\pi =\pm \frac{\pi }{6}+2\pi n,\; n\in Z\\\\\frac{x}{3}=\pm \frac{\pi }{6}+10\pi +2\pi n,\; n\in Z$

$x=\pm \frac{\pi }{2}+30\pi +6\pi n=\left [ {{\frac{61\pi }{2}+6\pi n,\; n\in Z} \atop {\frac{59\pi }{2}+6\pi n,\; n\in Z }} \right. \\\\3)\; \; cos(\frac{5}{6}\cdot x)=\frac{\sqrt3}{2}\\\\ \frac{5}{6}\cdot x=\pm \frac{\pi }{6}+2\pi n,\; n\in Z\\\\x=\pm \frac{\pi }{5}+ \frac{12\pi n}{5}\; ,\; n\in Z$