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$\sin8x\cos x=\sin 5x\cos 2x$ $\Rightarrow \dfrac{1}{2}(\sin 7x+\sin 9x)=\dfrac{1}{2}(\sin 3x+\sin 7x)$ $\Rightarrow \sin9x=\sin3x$ $\Rightarrow \left[\begin{array}{l}9x=3x+k2\pi \\ 9x=\pi-3x+k2pi \end{array} \right .\Rightarrow \left[\begin{array}{l}x=k\dfrac{\pi}{3} \\ x=\dfrac{\pi}{12}+k\dfrac{pi}{6} \end{array} \right .(k\in\mathbb Z)$
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