1) Xét $\cos x=0$ không là nghiệm của phương trình nên ta chia cả hai vế cho $\cos^3 x$ thì phương trình trở thành:
$\begin{array}{l} 2{\sin ^3}x + 4{\cos ^3}x = 3\sin x\\ \Leftrightarrow 2{\sin ^3}x + 4{\cos ^3}x - 3\sin x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0\\ \Leftrightarrow 2{\sin ^3}x + 4{\cos ^3}x - 3{\sin ^3}x - 3\sin x{\cos ^2}x = 0\\ \Leftrightarrow - {\sin ^3}x + 4{\cos ^3}x - 3\sin x{\cos ^2}x = 0\\ \Leftrightarrow - {\tan ^3}x - 3\tan x + 4 = 0\\ \Leftrightarrow {\tan ^3}x + 3\tan x - 4 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ \tan x = - \dfrac{4}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \arctan \left( { - \dfrac{4}{3}} \right) + k\pi \end{array} \right. \end{array}$
2)
$\begin{array}{l} {\sin ^2}3x - {\cos ^2}4x = {\sin ^2}5x - {\cos ^2}6x\\ \Leftrightarrow {\sin ^2}3x - {\sin ^2}5x = {\cos ^2}4x - {\cos ^2}6x\\ \Leftrightarrow \left( {\sin 3x - \sin 5x} \right)\left( {\sin 3x + \sin 5x} \right) = \left( {\cos 4x - \cos 6x} \right)\left( {\cos 4x + \cos 6x} \right)\\ \Leftrightarrow - 2\cos 4x\sin x.2\sin 4x\cos x = 2\sin 5x\sin x.2\cos 5x.\cos x\\ \Leftrightarrow - \sin 8x\sin 2x = \sin 10x\sin 2x\\ \Leftrightarrow \sin 2x\left( {\sin 10x + \sin 8x} \right) = 0\\ \Leftrightarrow 2\sin 2x\sin 9x\cos x = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{k\pi }}{2}\\ x = \dfrac{{k9\pi }}{2}\\ x = \dfrac{\pi }{2} + k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{k\pi }}{2}\\ x = \dfrac{{k9\pi }}{2} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
3) Điều kiện xác định $\cos 5x\ne 0\Rightarrow x\ne \dfrac{\pi}{10}+k\dfrac{\pi}{5}$. Khi đó phương trình đã cho trở thành
$\begin{array}{l} \cos 3x.\tan 5x = \sin 7x\\ \Leftrightarrow \cos 3x.\dfrac{{\sin 5x}}{{\cos 5x}} = \sin 7x\\ \Leftrightarrow \cos 3x\sin 5x = \sin 7x\cos 5x\\ \Leftrightarrow \dfrac{1}{2}\left( {\sin 8x + \sin 2x} \right) = \dfrac{1}{2}\left( {\sin 12x + \sin 2x} \right)\\ \Leftrightarrow \sin 8x = \sin 12x\\ \Leftrightarrow \left[ \begin{array}{l} 12x = 8x + k2\pi \\ 12x = \pi - 8x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{k\pi }}{2}\\ x = \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{{10}} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
