Đáp án:
$\left[\begin{array}{l}x = k2\pi\\ x = - \dfrac{2\pi}{3} +k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\cos\dfrac{\pi}{3}\cos x - \sin\dfrac{\pi}{3}\sin x = \dfrac{1}{2}\\ \Leftrightarrow \cos\left(\dfrac{\pi}{3} + x\right) = \cos\dfrac{\pi}{3}\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{\pi}{3} + x = \dfrac{\pi}{3} + k2\pi\\\dfrac{\pi}{3} + x = - \dfrac{\pi}{3} +k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k2\pi\\ x = - \dfrac{2\pi}{3} +k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
