Giải phương trình 2cos(2cosx)= √3

$2cos(2cosx) = \sqrt{3}$

$\Leftrightarrow cos(2cosx) = cos\dfrac{\pi}{6}$

$\Leftrightarrow 2cosx = \pm \dfrac{\pi}{6} + k2\pi$

$\Leftrightarrow cosx = \pm \dfrac{\pi}{12} + k\pi \, \, \, (k \in \Bbb Z)$

Do $|cosx| \leq 1$

nên $-1 \leq \pm \dfrac{\pi}{12} + k\pi \leq 1 \, \, \, (k \in \Bbb Z)$

$\Rightarrow k = 0$

$\Rightarrow \left[\begin{array}{l}x = \pm arccos\left(\dfrac{\pi}{12}\right) + l2\pi\\ x = \pm arccos\left(-\dfrac{\pi}{12}\right) + l2\pi\end{array}\right.\, \, (l \in \Bbb Z)$