Nghiệm của pt 2cos x + 1 = 0 trên đoạn [0;12pi]?

`2cos x + 1 = 0`

`-> cos x = -1/2`

`-> x = ±(2\pi)/3 + k2\pi` `(k in ZZ)`

`text{Ta có}`

`0 <= x <= 12pi`

`->` \(\left[ \begin{array}{l}0 ≤ \dfrac{2\pi}{3} + k2\pi ≤ 12\pi\\0 ≤ \dfrac{-2\pi}{3} + k2\pi ≤ 12\pi\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}\dfrac{-2\pi}{3} ≤ k2\pi ≤ \dfrac{34\pi}{3}\\\dfrac{2\pi}{3} ≤ k2\pi ≤ \dfrac{38\pi}{3}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}-\dfrac{1}{3} ≤ k ≤ \dfrac{17}{3}\\\dfrac{1}{3} ≤ k ≤ \dfrac{19}{3}\end{array} \right.\) 

`-> 1/3 <= k <= 17/3`

`-> k = {1; 2; 3; 4; 5}`