1. If the line $x \cos \theta+y \sin \theta=\cos ^{2} \theta-\sin ^{2} \theta$ $(0<\theta<\pi)$ intersects the circle $x^{2}+y^{2}=\frac{1}{4}$, then the range of values for $\theta$ is $\qquad$
二、1. $\frac{\pi}{6} \leqslant \theta \leqslant \frac{\pi}{3}$ or $\frac{2 \pi}{3} \leqslant \theta \leqslant \frac{5 \pi}{6}$.
Let the distance from the center of the circle to the line be
$$
d=\frac{\left|\cos ^{2} \theta-\sin ^{2} \theta\right|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=|\cos 2 \theta| \leqslant r=\frac{1}{2},
$$

i.e., $k \pi+\frac{\pi}{3} \leqslant 2 \theta \leqslant k \pi+\frac{2 \pi}{3}, k \in \mathbf{Z}$.
Since $0<\theta<\pi$, taking $k=0,1$ yields $\frac{\pi}{6} \leqslant \theta \leqslant \frac{\pi}{3}$ or $\frac{2 \pi}{3} \leqslant \theta \leqslant \frac{5 \pi}{6}$.