2. In $\triangle A B C$, the internal angles $\angle A, \angle B, \angle C$ are opposite to the sides $a, b, c$ respectively. If $a=2, b=3, \angle C=$ $2 \angle A$, then $\cos C=$ $\qquad$
2. $\frac{1}{4}$.

From the problem, we know
$$
\begin{array}{l}
\sin C=\sin 2 A=2 \sin A \cdot \cos A \\
\Rightarrow c=2 a \cos A=2 a \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c} \\
\quad=4 \times \frac{9+c^{2}-4}{6 c} \\
\Rightarrow c^{2}=10 \\
\Rightarrow \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{1}{4} .
\end{array}
$$