$\begin{array}{l}a) \,\,2\sin\left(3x - \dfrac{\pi}{4}\right) + 1 = 0\\ \Leftrightarrow \sin\left(3x - \dfrac{\pi}{4}\right) = - \dfrac{1}{2}\\ \Leftrightarrow \left[\begin{array}{l}3x - \dfrac{\pi}{4} = - \dfrac{\pi}{6} + k2\pi\\3x - \dfrac{\pi}{4} = \dfrac{7\pi}{6} +k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x =\dfrac{\pi}{12} + k2\pi\\3x = \dfrac{17\pi}{12} +k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{36} + k\dfrac{2\pi}{3}\\x = \dfrac{17\pi}{36} +k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\,2\cos2x - 4\cos x - 1 = 0\\ \Leftrightarrow 2(2\cos^2x - 1) - 4\cos x - 1 = 0\\ \Leftrightarrow 4\cos^2x - 4\cos x - 3 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos x = - \dfrac{1}{2}\\\cos x = \dfrac{3}{2}\quad (loại)\end{array}\right.\\ \Leftrightarrow x = \pm \dfrac{2\pi}{3} +k2\pi \quad (k \in \Bbb Z)\\ c)\,\,\sin^3x\cos x - \cos^3x\sin x = \dfrac{1}{4}\\ \Leftrightarrow 4\sin x\cos x(\sin^2x - \cos^2x) = 1\\ \Leftrightarrow -2\sin2x.\cos2x = 1\\ \Leftrightarrow \cos4x = - 1\\ \Leftrightarrow 4x = \pi + k2\pi\\ \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\quad (k \in \Bbb Z) \end{array}$
