c) $2cos2x = \sqrt{6}(cosx - sinx)$

$\Leftrightarrow 2(cos^2x - sin^2x) = \sqrt{6}(cosx - sinx)$

$\Leftrightarrow 2(cosx - sinx)(cosx + sinx) - \sqrt{6}(cosx - sinx) = 0$

$\Leftrightarrow (cosx - sinx)(2cosx + 2sinx - \sqrt{6}) = 0$

Với $cosx - sinx = 0$

$\Leftrightarrow sinx = cosx$

$\Leftrightarrow tanx = 1$

$\Leftrightarrow x = \dfrac{\pi}{4} + k\pi$

Với $2cosx + 2sinx = \sqrt{6}$

$\Leftrightarrow \dfrac{\sqrt{2}}{2}cosx + \dfrac{\sqrt{2}}{2}sinx = \dfrac{\sqrt{3}}{2}$

$\Leftrightarrow sin\dfrac{\pi}{4}.cosx+ sinx.cos\dfrac{\pi}{4} = sin\dfrac{\pi}{3}$

$\Leftrightarrow sin(x + \dfrac{\pi}{4}) = sin\dfrac{\pi}{3}$

$\Leftrightarrow x + \dfrac{\pi}{4} = \dfrac{\pi}{3} + k2\pi \, hoặc \, x + \dfrac{\pi}{4} = \pi - \dfrac{\pi}{3} + k2\pi$

$\Leftrightarrow x = \dfrac{\pi}{12} + k2\pi \, hoặc \, x = \dfrac{7\pi}{12} + k2\pi \, (k \in \Bbb Z)$

Vậy phương trình có các nghiệm $x = \left\{\dfrac{\pi}{4} + k\pi; \dfrac{\pi}{12} + k2\pi; \dfrac{7\pi}{12} + k2\pi \right\} \, (k \in \Bbb Z)$

d) $3sinx = 3 - \sqrt{3}cosx$

$\Leftrightarrow \sqrt{3}sinx + cosx = \sqrt{3}$

$\Leftrightarrow \dfrac{\sqrt{3}}{2}sinx + \dfrac{1}{2}cosx = \dfrac{\sqrt{3}}{2}$

$\Leftrightarrow sinx.cos\dfrac{\pi}{6} - sin\dfrac{\pi}{6}.cosx = sin\dfrac{\pi}{3}$

$\Leftrightarrow sin(x - \dfrac{\pi}{6}) = sin\dfrac{\pi}{3}$

$\Leftrightarrow x - \dfrac{\pi}{6} = \dfrac{\pi}{3} + k2\pi \, hoặc \, x - \dfrac{\pi}{6} = \pi - \dfrac{\pi}{3} + k2\pi$

$\Leftrightarrow x = \dfrac{\pi}{2} + k2\pi \, hoặc \, x = \dfrac{5\pi}{6} + k2\pi \, (k \in \Bbb Z)$

Vậy phương trình có nghiệm $x = \left\{\dfrac{\pi}{2} + k2\pi; \dfrac{5\pi}{6}+ k2\pi \right\} \, (k \in \Bbb Z)$