2. In $\triangle A B C$, the internal angles $A, B, C$ are opposite the sides $a, b, c$ respectively. If $a=2, b=3, C=2 A$, then $\cos C=$
Given $\frac{1}{4}$.
Analysis From $C=2A$, we know $\sin C=\sin 2A=2 \sin A \cos A$. Combining with the Law of Sines, we get $c=2a \cos A$. From $a=2, b=3$, and the Law of Cosines, we have
$$
c=2a \cos A=2a \cdot \frac{b^{2}+c^{2}-a^{2}}{2bc}, c=4 \times \frac{9+c^{2}-4}{6c}
$$

So $c^{2}=10, c=\sqrt{10}$, thus
$$
\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{4+9-10}{2 \cdot 2 \cdot 3}=\frac{1}{4}
$$