2. In $\triangle A B C$, if $\sin A+\cos A=-\frac{1}{3}$, then $\cos 2 A=$ . $\qquad$
2. $\frac{\sqrt{17}}{9}$.

From $\sin A+\cos A=-\frac{1}{3}$, we know
$$
1+\sin 2 A=\frac{1}{9} \Rightarrow \sin 2 A=-\frac{8}{9} \text {. }
$$

From $0<\angle A<\pi, \sin A+\cos A=-\frac{1}{3}$, we know
$$
\frac{3 \pi}{4}<\angle A<\pi \text {. }
$$

Therefore, $\cos 2 A=\sqrt{1-\left(-\frac{8}{9}\right)^{2}}=\frac{\sqrt{17}}{9}$.