Đáp án:
$\left[\begin{array}{l}x= - \dfrac{\pi}{6}+ k2\pi\\x= \dfrac{\pi}{24234} + k\dfrac{2\pi}{4039}\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sqrt3\cos2019x + \sin2019x = 2\cos2020x\\ \Leftrightarrow \dfrac{\sqrt3}{2}\cos2019x + \dfrac{1}{2}\sin2019x = \cos2020x\\ \Leftrightarrow \cos\dfrac{\pi}{6}\cos2019x + \sin\dfrac{\pi}{6}\sin2019x = \cos2020x\\ \Leftrightarrow \cos\left(2019x - \dfrac{\pi}{6}\right) = \cos2020x\\ \Leftrightarrow \left[\begin{array}{l}2019x - \dfrac{\pi}{6} = 2020x + k2\pi\\2019x - \dfrac{\pi}{6} = - 2020x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x= - \dfrac{\pi}{6}+ k2\pi\\x= \dfrac{\pi}{24234} + k\dfrac{2\pi}{4039}\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
