$9.276 \sin 2 x \cdot \sin 3 x-\cos 2 x \cdot \cos 3 x>\sin 10 x$.
Solution. $-\cos (2 x+3 x)>\sin 10 x \Leftrightarrow \cos 5 x+\sin 10 x0, \\ \sin 5 x-\frac{1}{2}\end{array} \Leftrightarrow\right.$

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$\Leftrightarrow \frac{\pi}{2}+2 \pi n<5 x<\frac{7 \pi}{6}+2 \pi n$.

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Answer: $\quad x \in\left(-\frac{\pi}{10}+\frac{2 \pi n}{5} ;-\frac{\pi}{30}+\frac{2 \pi n}{5}\right) \cup\left(\frac{\pi}{10}+\frac{2 \pi n}{5} ; \frac{7 \pi}{30}+\frac{2 \pi n}{5}\right)$.