1. Let $-1<a<0, \theta=\arcsin a$, then the solution set of the inequality $\sin x<a$ is
A. $\{x \mid 2 n \pi+\theta<x<(2 n+1) \pi-\theta, n \in \mathbf{Z}\}$
B. $\{x \mid 2 n \pi-\theta<x<(2 n+1) \pi+\theta, n \in \mathbf{Z}\}$
C. $\{x \mid(2 n-1) \pi+\theta<x<2 n \pi-\theta, n \in \mathbf{Z}\}$
D. $\{x \mid(2 n-1) \pi-\theta<x<2 n \pi+\theta, n \in \mathbf{Z}\}$
Solve:
As shown in the figure, draw the line $y=a$ to intersect the unit circle at two points, then the shaded area (excluding the boundary) is the solution set that satisfies the condition. Within one period, it can be expressed as $\{x \mid-\pi-\theta<x<\theta\}$. Therefore, the solution set of the inequality $\sin x<a$ is $\{x \mid(2 n-1) \pi-\theta<x<2 n \pi+\theta, n \in \mathbf{Z}\}$, so the answer is $D$.