Đáp án:
$\left[\begin{array}{l}x +\dfrac{\pi}{6} =\dfrac{\pi}{12} + k\pi\\x +\dfrac{\pi}{6} = -\dfrac{5\pi}{12} + k\pi\end{array}\right.(k\in\Bbb Z)$
Giải thích các bước giải:
$\sqrt3\cos2x -\sin2x =\sqrt2$
$\Leftrightarrow \dfrac{\sqrt3}{2}\cos2x -\dfrac{1}{2}\sin2x =\dfrac{\sqrt2}{2}$
$\Leftrightarrow \cos\left(2x +\dfrac{\pi}{6}\right) =\cos\dfrac{\pi}{4}$
$\Leftrightarrow\left[\begin{array}{l}2x +\dfrac{\pi}{6} =\dfrac{\pi}{4} + k2\pi\\2x +\dfrac{\pi}{6} = -\dfrac{\pi}{4} + k2\pi\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x +\dfrac{\pi}{6} =\dfrac{\pi}{12} + k\pi\\x +\dfrac{\pi}{6} = -\dfrac{5\pi}{12} + k\pi\end{array}\right.(k\in\Bbb Z)$
