9. The solution set of the inequality $\sin 2 \pi x|\sin 2 \pi x|>\cos 2 \pi x|\cos 2 \pi x|$ is $\qquad$
Answer
Since $x|x|=\left\{\begin{array}{l}-x^{2}, x<0 \\ x^{2}, x \geq 0\end{array}\right.$, we have $\sin 2 \pi x|\sin 2 \pi x|>\cos 2 \pi x|\cos 2 \pi x| \Leftrightarrow \sin^2 2 \pi x > \cos^2 2 \pi x \Leftrightarrow \cos 4 \pi x < 0 \Leftrightarrow x \in\left[k+\frac{1}{8}, k+\frac{5}{8}\right](k \in \mathbf{Z}) .
$
10. Given $x>0$, and $x,[x], x-[x]$ form a geometric sequence (where $[x]$ represents the greatest integer not exceeding $x$), then $x=$ $\qquad$ solution
$$
[x]^{2}=x(x-[x]) \Rightarrow x^{2}-[x] x-[x]^{2}=0,
$$

When $[x]=1$, $x^{2}-x-1=0 \Rightarrow x=\frac{\sqrt{5}+1}{2}$.
When $[x] \geqslant 2$, $x=\frac{\sqrt{5}+1}{2}[x]<[x]+1 \Rightarrow[x]<\frac{\sqrt{5}+1}{2}$, which is a contradiction.