Side 5. If $\theta$ is an angle in the second quadrant; and satisfies $\cos \frac{\theta}{3}-\sin \frac{\theta}{2}=\sqrt{1}-\sin \theta$, then $\frac{\theta}{2}$
(A) is an angle in the first quadrant;
(B) is an angle in the third quadrant;
(C) may be an angle in the first quadrant; or may be an angle in the third quadrant; $\square$
(D) is an angle in the second quadrant.
Among the four answers, since $\theta$ is in the second quadrant, it is known that $\frac{\theta}{2}$ cannot be in the second quadrant, so (D) is eliminated; also, because $\cos \frac{\theta}{2} - \sin \frac{\theta}{2} = \sqrt{1 - \sin \theta} \geqslant 0$, it follows that $\cos \frac{\theta}{2} \geqslant \sin \frac{\theta}{2}$. Since $\theta$ is in the second quadrant, the terminal side of $\frac{\theta}{2}$ must lie within $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ or $\left(\frac{5}{4} \pi, \frac{3}{2} \pi\right)$. By the monotonicity of the cosine function, $\frac{\theta}{2}$ cannot lie within $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, thus (A) and (C) are eliminated; therefore, (B) is the correct answer.

Above, we used the method of elimination to solve the multiple-choice question relatively quickly and correctly, which is a good approach. Readers who are interested can give it a try.