8. When $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, the function $f(x)$ satisfies
$2 f(-\sin x)+3 f(\sin x)=\sin 2 x$.
Then $f(x)$ is ( ).
(A) odd function
(B) neither odd nor even function
(C) even function
(D) both odd and even function
8. A.

Let $t=\sin x$, then $\sin 2 x=2 t \sqrt{1-t^{2}}$, so
$$
2 f(-t)+3 f(t)=2 t \sqrt{1-t^{2}}.
$$

Thus, $2 f(t)+3 f(-t)=-2 t \sqrt{1-t^{2}}$.
(1) + (2) gives
$$
5(f(-t)+f(t))=0 \quad (-1 \leqslant t \leqslant 1).
$$

Therefore, $f(-t)=-f(t)$.
Hence, $f(x)$ is an odd function on $[-1,1]$.