Ta có: \(4\sin 3x+\sin 5x-2\sin x\cos 2x=0\)
\(\Leftrightarrow 4\sin 3x+\sin 5x-\)
\(2\dfrac{1}{2}\left[ {\sin (x - 2x) + \sin (x + 2x)} \right]=0\)
\(\Leftrightarrow 4\sin 3x+\sin 5x-\)
\(\left[ {\sin (- x) + \sin 3x} \right]=0\)
\(\Leftrightarrow 3\sin 3x+\sin 5x+\sin x=0\)
\(\Leftrightarrow 3\sin 3x+\)
\(2\sin\dfrac{{5x + x}}{2}\cos \dfrac{{5x - x}}{2}=0\)
\(\Leftrightarrow 3\sin 3x+2\sin 3x\cos 2x=0\)
\(\Leftrightarrow \sin 3x(3+2\cos 2x)=0\)
\(\Leftrightarrow\left[ \begin{array}{l} \sin 3x = 0\\\cos 2x=-\dfrac{3}{2}<-1\text{(loại)}\end{array} \right. \)
\(\sin 3x=0\Leftrightarrow 3x = k\pi,k\in\mathbb{Z}\)
\(x=k\dfrac{\pi}{3},k\in\mathbb{Z}\)
Vậy phương trình có nghiệm là \(x=k\dfrac{\pi}{3},k\in\mathbb{Z}\).
