Đáp án:
$\left[\begin{array}{l}x=-\dfrac{\pi}{24} + k\pi\\x=-\dfrac{5\pi}{24} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$2\cos2x -2\sin2x =\sqrt2$
$\to \dfrac{\sqrt2}{2}\cos2x -\dfrac{\sqrt2}{2}\sin2x =\dfrac12$
$\to \cos\left(2x +\dfrac{\pi}{4}\right)=\cos\dfrac{\pi}{3}$
$\to \left[\begin{array}{l}2x +\dfrac{\pi}{4}=\dfrac{\pi}{6} + k2\pi\\2x +\dfrac{\pi}{4} =-\dfrac{\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x=-\dfrac{\pi}{24} + k\pi\\x=-\dfrac{5\pi}{24} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
