Giải:
\[5sin3x = 3sin5x \Leftrighrarrow 2sin3x = 3(sin5x - sin 3x)\]
\[\Leftrightarrow 2(sinx - 4sin^3x) = 6cos4x.sinx <=> (3-4sin^2x - 3cos4x).sinx = 0\]
\[\Leftrightarrow [3 - 2(1-cos2x) - 3(2cos^22x -1)].sinx = 0\]
\[\Leftrightarrow (3cos^22x - cos2x -2).sinx = 0\]
\[\Leftrightarrow \left[ \begin{matrix}cos 2x = 1 \\ cos2x = \frac{-2}{3} \\ sinx = 0\end{matrix}\right.\]
\[\Leftrightarrow \left[ \begin{matrix}cos2x = \frac{-2}{3} = cos2\alpha \\ sinx = 0\end{matrix}\right.\]
\[\Leftrightarrow \left[ \begin{matrix}2x = \pm 2\alpha + k2\pi \\ x = k\pi \end{matrix}\right.\]
\[\Leftrightarrow \left[ \begin{matrix}x = \pm \alpha + k2\pi \\ x =k\pi\end{matrix}\right.\]
P/S. thầy ng` điên sửa giùm e chõ dễ đọc nha
