Giải phương trình:

a) sin 5x + sin 8x + sin 3x = 0;

b) \(4{\cos ^3}x + 3\sqrt 2 \sin 2x = 8\cos x\).

a) sin 5x + sin 8x + sin 3x = 0

Û 2sin 4x.cos x + 2sin 4x.cos 4x = 0

Û 2sin 4x(cos x + cos 4x) = 0

\( \Leftrightarrow 4\sin 4x\,.\,\cos \frac{{5x}}{2}\cos \frac{{3x}}{2} = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\sin 4x = 0\\\cos \frac{{5x}}{2} = 0\\\cos \frac{{3x}}{2} = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}4x = k\pi \\\frac{{5x}}{2} = \frac{\pi }{2} + k\pi \\\frac{{3x}}{2} = \frac{\pi }{2} + k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{4}\\x = \frac{\pi }{5} + \frac{{k2\pi }}{5}\\x = \frac{\pi }{3} + \frac{{k2\pi }}{3}\end{array} \right.\)

Vậy \(x = \frac{{k\pi }}{4},\;x = \frac{\pi }{5} + \frac{{k2\pi }}{5},\;x = \frac{\pi }{3} + \frac{{k2\pi }}{3}\;\left( {k \in \mathbb{Z}} \right)\)

b) \(4{\cos ^3}x + 3\sqrt 2 \sin 2x = 8\cos x\)

\( \Leftrightarrow 2\cos x\left( {2{{\cos }^2}x + 3\sqrt 2 \sin x - 4} \right) = 0\)

\( \Leftrightarrow 2\cos x\left( { - 2{{\sin }^2}x + 3\sqrt 2 \sin x - 2} \right) = 0\)

Vậy \(x = \frac{\pi }{2} + k\pi ,\;x = \frac{\pi }{4} + k2\pi ,\;x = \frac{{3\pi }}{4} + k2\pi \;\left( {k \in \mathbb{Z}} \right)\)