Lời giải của GV Vungoi.vn
TXĐ: \[D = \mathbb{R}\].
\[\begin{array}{l}\cos \left[ {x - \dfrac{\pi }{3}} \right] = \cos \left[ {2x + \dfrac{\pi }{6}} \right]\\ \Leftrightarrow \left[ \begin{array}{l}2x + \dfrac{\pi }{6} = x - \dfrac{\pi }{3} + k2\pi \\2x + \dfrac{\pi }{6} = - x + \dfrac{\pi }{3} + k2\pi \end{array} \right.\left[ {k \in \mathbb{Z}} \right]\\ \Leftrightarrow \left[ \begin{array}{l}x = - \dfrac{\pi }{2} + k2\pi \\x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\end{array} \right.\,\,\,\left[ {k \in \mathbb{Z}} \right]\end{array}\]
Xét họ nghiệm \[x = - \dfrac{\pi }{2} + k2\pi \,\,\,\left[ {k \in \mathbb{Z}} \right]\], cho \[x \in \left[ { - \pi ;\pi } \right]\].
\[\begin{array}{l} \Rightarrow - \pi