Giúp mk câu này vs mn

`2cos(2x- \pi/3)- \sqrt{3} =0`

`<=>cos(2x- \pi/3) = {\sqrt{3}} /2`

`<=>` $\left[ \begin{array}{l}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\dfrac{-\pi}{6}+k2\pi\end{array} \right.$ 

`<=>` $\left[ \begin{array}{l}2x=\dfrac{\pi}{2}+k2\pi\\2x=\dfrac{\pi}{6}+k2\pi\end{array} \right.$ 

`<=>` $\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{array} \right.$ 

`+)` `-2\pi < \frac{\pi}{4}+k\pi < 3\pi`

`<=>{-9\pi}/4 < k\pi < {11\pi}/4`

`<=>` `{-9}/4 < k < {11}/4`

`=>k={-2;-1;0;1;2}`

`=>` có `5` nghiệm

`+)` `-2\pi < \frac{\pi}{12}+k\pi < 3\pi`

`<=>` `{-25\pi}/{12} < k\pi < {35\pi}/{12}`

`<=>` `{-25}/{12} < k < {35}/{12}`

`=>k={-2;-1;0;1;2}`

`=>` có `5` nghiệm

Vậy,  phương trình có tổng cộng `5+5=10` nghiệm