11. Given that $\theta$ is an acute angle, and $\frac{\cos 3 \theta}{\cos \theta}=\frac{1}{3}$. Then $\frac{\sin 3 \theta}{\sin \theta}=$ $\qquad$
$=.11 \cdot \frac{7}{3}$.
Solution 1: From the given and the triple angle cosine formula, we get $4 \cos ^{2} \theta-3=\frac{1}{3}$, which means $4 \cos ^{2} \theta=\frac{10}{3}$.
Therefore, $\frac{\sin 3 \theta}{\sin \theta}=3-4 \sin ^{2} \theta=4 \cos ^{2} \theta-1=\frac{7}{3}$.
Solution 2: Let $\frac{\sin 3 \theta}{\sin \theta}=x$, then
$$
\begin{array}{l}
x-\frac{1}{3}=\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta} \\
=\frac{\sin (3 \theta-\theta)}{\sin \theta \cdot \cos \theta}=\frac{\sin 2 \theta}{\frac{1}{2} \sin 2 \theta}=2 .
\end{array}
$$

Thus, $x=2+\frac{1}{3}=\frac{7}{3}$.