New proposition: Let $x \in\left(0, \frac{\pi}{2}\right)$ and $\frac{1}{\sin x}=\frac{1}{\sin 2 x}+\frac{1}{\sin 4 x}+\frac{1}{\sin 8 x}$, find the value of $x$.


Since $\frac{1}{\sin 2 x}=\cot x-\cot 2 x$, then
$$
\frac{1}{\sin x}=(\cot x-\cot 2 x)+(\cot 2 x-\cot 4 x)+(\cot 4 x-\cot 8 x)=\cot x-\cot 8 x \text {, }
$$

thus $\frac{1}{\sin x}-\cot x=-\cot 8 x \Leftrightarrow \frac{1-\cos x}{\sin x}=-\cot 8 x \Leftrightarrow \tan \frac{x}{2}=-\cot 8 x$
$$
\begin{array}{l}
\Leftrightarrow \cot \left(90^{\circ}+\frac{x}{2}\right)=\cot 8 x \Leftrightarrow 8 x=90^{\circ}+\frac{x}{2}+180^{\circ} \cdot k, k \in \mathbf{Z} \\
\Leftrightarrow x=12^{\circ}+24^{\circ} \cdot k, k \in \mathbf{Z} .
\end{array}
$$

Since $x \in\left(0,90^{\circ}\right)$, we have $x=12^{\circ}$ or $36^{\circ}$ or $60^{\circ}$ or $84^{\circ}$.