Question

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Answer 1

$\displaystyle\bf\\1)\\\\tgx=\frac{\sqrt{3} }{3} \\\\x=arctg\frac{\sqrt{3} }{3} +\pi n,n\in Z\\\\\boxed{x=\frac{\pi }{6} +\pi n,n\in Z}\\\\\\2)\\\\Cos\Big(\frac{\pi }{6} -\frac{x}{3} \Big)=-1\\\\\\\frac{\pi }{6} -\frac{x}{3} =\pi +2\pi n,n\in Z\\\\\\-\frac{x}{3} =\pi -\frac{\pi }{6} +2\pi n,n\in Z\\\\\\-\frac{x}{3}=\frac{5\pi }{6} +2\pi n,n\in Z\\\\\\x=-\frac{5\pi }{2} +6\pi n,n\in Z$