$\begin{array}{l}a)\,\,4\cos^2x + 8\cos x + 3 = 0\\ \Leftrightarrow (2\cos x + 1)(2\cos x + 3) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos x = - \dfrac{1}{2}\\\cos x = - \dfrac{3}{2}\quad (loại)\end{array}\right.\\ \Leftrightarrow x = \pm \dfrac{2\pi}{3} + k2\pi\quad (k \in \Bbb Z)\\ b)\,\,2\sin3x - 2\cos3x = 2\\ \Leftrightarrow \sin3x - \cos3x = 1\\ \Leftrightarrow \sqrt2\sin\left(3x - \dfrac{\pi}{4}\right) = 1\\ \Leftrightarrow \sin\left(3x - \dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}3x-\dfrac{\pi}{4} = \dfrac{\pi}{4} +k2\pi\\3x - \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x= \dfrac{\pi}{2} +k2\pi\\3x=\pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x= \dfrac{\pi}{6} +k\dfrac{2\pi}{3}\\x=\dfrac{\pi}{3} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
