[tex]Wiemy,~~ze:\\
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cos\alpha =-\dfrac{\sqrt{3} }{2} ~~\Rightarrow~~(~~\alpha =\dfrac{5}{6} \pi +2k\pi ~~\lor~~\alpha =\dfrac{7}{6} \pi +2k\pi~~)~~\land~~k\in C\\
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cos(2x-\dfrac{1}{6} \pi )=-\dfrac{\sqrt{3} }{2}\\
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2x-\dfrac{1}{6} \pi=\dfrac{5}{6} \pi +2k\pi ~~\lor ~~2x-\dfrac{1}{6} \pi=\dfrac{7}{6} \pi +2k\pi ~~~~\land ~~k\in C\\
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2x=\dfrac{5}{6} \pi+\dfrac{1}{6} \pi +2k\pi ~~\lor~~2x=\dfrac{7}{6} \pi+\dfrac{1}{6} \pi +2k\pi~~~~\land ~~k\in C\\
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[tex]2x=\pi +2k\pi ~~\mid \div 2~~\lor~~2x=\dfrac{4}{3} \pi +2k\pi ~~\mid \div 2~~~~\land ~~k\in C\\
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x=\dfrac{1}{2} \pi +k\pi ~~\lor~~x=\dfrac{2}{3} \pi +k\pi ~~~~~~~~\land ~~k\in C[/tex]
