10. (2007 Zhejiang Province High School Mathematics Competition A Paper) Let $f(x)$ be an odd function defined on $\mathbf{R}$, and satisfies $f(x+2)=-f(x)$; also, when $0 \leqslant x \leqslant 1$, $f(x)=\frac{1}{2} x$, then $\left\{x \left\lvert\, f(x)=-\frac{1}{2}\right.\right\}=$ $\qquad$ .
10. According to the problem, $f(x+4)=-f(x+2)=f(x)$, which means $f(x)$ is a periodic function with a period of 4.

When $0 \leqslant x \leqslant 1$, $f(x)=\frac{1}{2} x$, and $f(x)$ is an odd function. Therefore, when $-1 \leqslant x<0$, $f(x)=\frac{1}{2} x$. Thus, we have $f(x)=\left\{\begin{array}{ll}\frac{1}{2} x, & -1 \leqslant x \leqslant 1, \\ -\frac{1}{2} x+1, & 1 \leqslant x \leqslant 3\end{array}\right.$. It follows that $f(-1)=f(3)=-\frac{1}{2}$. Since $f(x)$ is a periodic function with a period of 4, we also have $f(4 k-1)=-\frac{1}{2},(k \in \mathbf{Z})$, so the answer is $\{x \mid x=4 k-1,(k \in \mathbf{Z})\}$.