3. Given that $\alpha, \beta$ satisfy
$$
\csc (\alpha-2 \beta), \csc \alpha, \csc (\alpha+2 \beta)
$$

form an arithmetic sequence with a non-zero common difference. Then the value of $\frac{\sin \alpha}{\cos \beta}$ could be ( ).
(A) $\pm 1$
(B) $\pm \sqrt{2}$
(C) $\pm \sqrt{3}$
(D) None of the above
3. B.

From the given, we have $\frac{2}{\sin \alpha}=\frac{1}{\sin (\alpha-2 \beta)}+\frac{1}{\sin (\alpha+2 \beta)}$.
Thus, $2 \sin (\alpha-2 \beta) \cdot \sin (\alpha+2 \beta)$
$$
=[\sin (\alpha-2 \beta)+\sin (\alpha+2 \beta)] \sin \alpha \text {, }
$$

which means $2 \sin (\alpha-2 \beta) \cdot \sin (\alpha+2 \beta)=2 \sin ^{2} \alpha \cdot \cos 2 \beta$.
The left side of the above equation $=\cos 4 \beta-\cos 2 \alpha$, so
$$
\begin{array}{l}
\cos ^{2} 2 \beta-\cos ^{2} \alpha=\sin ^{2} \alpha \cdot \cos 2 \beta \\
=\sin ^{2} \alpha-2 \sin ^{2} \beta \cdot \sin ^{2} \alpha,
\end{array}
$$

which implies $\sin ^{2} 2 \beta=2 \sin ^{2} \beta \cdot \sin ^{2} \alpha$.
If $\sin \beta=0$, then the common difference of the original arithmetic sequence is 0, so $\sin \beta \neq 0$, and we have $2 \cos ^{2} \beta=\sin ^{2} \alpha$. Therefore, $\frac{\sin \alpha}{\cos \beta}= \pm \sqrt{2}$ and it can be achieved.