1. Let $-1<a<0, \theta=\arcsin a$, then the solution set of the inequality $\sin x<a$ is
(A) $\{x \mid 2 n \pi+\theta<x<(2 n+1) \pi-\theta, n \in Z\}$;
(B) $\{x \mid 2 n \pi-\theta<x<(2 n+1) \pi+\theta, n \in Z\}$;
(C) $\{x \mid(2 n-1) \pi+\theta<x<2 n \pi-\theta, n \in Z\}$;
(D) $\{x \mid(2 n-1) \pi-\theta<x<2 n \pi+\theta, n \in Z\}$.
1. (D)
$$
\because \quad-1<a<0, \quad \because \theta=\arcsin a \in\left(-\frac{\pi}{2}, 0\right) \text {. }
$$

In the interval $(-\pi, \pi)$, the solutions to the equation $\sin x=a$ are $x=-\pi-\arcsin a$ or $x=\arcsin a$. Therefore, the solution set of the inequality $\sin x<a$ is (D).