Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\\x = \pm \dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\cos x + \cos2x + \cos3x = \sin x + \sin2x + \sin3x\\ \Leftrightarrow (\cos x - \sin x) + (\cos3x - \sin3x) + (\cos2x - \sin2x) = 0\\ \Leftrightarrow \cos\left(x + \dfrac{\pi}{4}\right) + \cos\left(3x + \dfrac{\pi}{4}\right) + \cos\left(2x + \dfrac{\pi}{4}\right) = 0\\ \Leftrightarrow 2\cos\left(2x + \dfrac{\pi}{4}\right).\cos x + \cos\left(2x + \dfrac{\pi}{4}\right) = 0\\ \Leftrightarrow \cos\left(2x + \dfrac{\pi}{4}\right).(2\cos x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos\left(2x + \dfrac{\pi}{4}\right) = 0\\2\cos x +1 =0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{4} = \dfrac{\pi}{2} + k\pi\\\cos x = - \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\\x = \pm \dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
