$9.2872 \sin ^{2} x-\sin x+\sin 3 x<1$.

Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.

$9.2872 \sin ^{2} x-\sin x+\sin 3 x<1$.
Solution.

$$
\begin{aligned}
& 2 \sin ^{2} x-\sin x+\sin 3 x-4 \sin ^{3} x=0 \Leftrightarrow \\
& \Leftrightarrow\left(2 \sin ^{2} x-1\right)(2 \sin x-1)>0 \Leftrightarrow\left(\sin x-\frac{1}{2}\right)\left(\sin x-\frac{\sqrt{2}}{2}\right)\left(\sin x+\frac{\sqrt{2}}{2}\right)>0 \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array}{l}
\sin x>\frac{\sqrt{2}}{2} \\
-\frac{\sqrt{2}}{2}<\sin x<\frac{1}{2}
\end{array}\right.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-218.jpg?height=389&width=467&top_left_y=479&top_left_x=366)

Answer: $x \in\left(\frac{\pi}{4}+2 \pi n ; \frac{3 \pi}{4}+2 \pi n\right) \cup\left(-\frac{\pi}{4}+2 \pi n ; \frac{\pi}{6}+2 \pi n\right) \cup\left(\frac{5 \pi}{6}+2 \pi n ; \frac{5 \pi}{4}+2 \pi n\right)$.