$9.2894 \sin x \cdot \sin 2 x \cdot \sin 3 x>\sin 4 x$.
## Solution.

$2 \sin x \cdot \sin 2 x \cdot \sin 3 x>\sin 2 x \cdot \cos 2 x \Leftrightarrow \sin 2 x(2 \sin x \cdot \sin 3 x-\cos 2 x)>0 \Leftrightarrow$ $\Leftrightarrow \sin 2 x(\cos 2 x-\cos 4 x-\cos 2 x)>0 \Leftrightarrow-\sin 2 x \cdot \cos 4 x>0 \Leftrightarrow$
$\Leftrightarrow\left(2 \sin ^{2} 2 x-1\right) \sin 2 x>0 \Leftrightarrow\left(\sin 2 x-\frac{1}{\sqrt{2}}\right)\left(\sin 2 x+\frac{1}{\sqrt{2}}\right) \sin 2 x>0 \Leftrightarrow$

$\Leftrightarrow\left[\begin{array}{l}\sin 2 x>\frac{1}{\sqrt{2}}, \\ -\frac{1}{\sqrt{2}}<\sin 2 x<0\end{array}\right.$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-219.jpg?height=357&width=444&top_left_y=165&top_left_x=486)

$\Leftrightarrow\left[\begin{array}{l}\frac{\pi}{4}+2 \pi n<2 x<\frac{3 \pi}{4}+2 \pi n \\ -\frac{\pi}{4}+2 \pi n<2 x<2 \pi n \\ \pi+2 \pi n<2 x<\frac{5 \pi}{4}+2 \pi n\end{array}\right.$

Answer: $\quad x \in\left(-\frac{\pi}{8}+\pi n ; \pi n\right) \cup\left(\frac{\pi}{8}+\pi n ; \frac{3 \pi}{8}+\pi n\right) \cup\left(\frac{\pi}{2}+\pi n ; \frac{5 \pi}{8}+\pi n\right)$.