Đáp án:

Giải thích các bước giải:

Đáp án:

$\left[\begin{array}{l}x = \pm \dfrac13\arccos\left(\dfrac{1 - \sqrt{97}}{12}\right) + k\dfrac{2\pi}{3}\\x = \pm \dfrac13\arccos\left(\dfrac{1 + \sqrt{97}}{12}\right) + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$\begin{array}{l}\quad 3\cos6x - \cos3x - 1 =0\\ \Leftrightarrow 3(2\cos^23x - 1) - \cos3x - 1 = 0\\ \Leftrightarrow 6\cos^23x - \cos3x - 4 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos3x = \dfrac{1 - \sqrt{97}}{12}\\\cos3x = \dfrac{1+\sqrt{97}}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x = \pm \arccos\left(\dfrac{1 - \sqrt{97}}{12}\right) + k2\pi\\3x = \pm \arccos\left(\dfrac{1 + \sqrt{97}}{12}\right) + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \pm \dfrac13\arccos\left(\dfrac{1 - \sqrt{97}}{12}\right) + k\dfrac{2\pi}{3}\\x = \pm \dfrac13\arccos\left(\dfrac{1 + \sqrt{97}}{12}\right) + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z) \end{array}$