a) $\sqrt3\sin2x + \cos2x = \sqrt2\sin x - \sqrt2\cos x$
$\Leftrightarrow\dfrac{\sqrt3}{2}\sin2x + \dfrac{1}{2}\cos2x = \dfrac{\sqrt2}{2}\sin x - \dfrac{\sqrt2}{2}\cos x$
$\Leftrightarrow \sin\left(2x + \dfrac{\pi}{6}\right) = \sin\left(x - \dfrac{\pi}{4}\right)$
$\Leftrightarrow\left[\begin{array}{l}2x + \dfrac{\pi}{6} = x - \dfrac{\pi}{4} + k2\pi\\2x + \dfrac{\pi}{6} = \dfrac{5\pi}{4} - x + k2\pi\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x = - \dfrac{5\pi}{12} + k2\pi\\x = \dfrac{13\pi}{36} + k\dfrac{2\pi}{3}\end{array}\right.$
b) $\cos x - \sqrt3\sin x = 2\cos\left(\dfrac{\pi}{3} - x \right)$
$\Leftrightarrow \dfrac{1}{2}\cos x - \dfrac{\sqrt3}{2}\sin x = \cos\left(\dfrac{\pi}{3} - x \right)$
$\Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{3} - x \right)$
$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{3} = \dfrac{\pi}{3} - x + k2\pi\\x + \dfrac{\pi}{3} = x - \dfrac{\pi}{3} +k2\pi\end{array}\right.$
$\Leftrightarrow x = k\pi \quad (k\in\Bbb Z)$
