2. In an acute triangle, the interior angles $A, B$ satisfy $\tan A-\frac{1}{\sin 2 A}=\tan B$ and $\cos ^{2} \frac{B}{2}=\frac{\sqrt{6}}{3}$, then $\sin 2 A=$
$$
\begin{array}{l}
\tan A-\frac{1}{\sin 2 A}=\tan A-\frac{\tan ^{2} A+1}{2 \tan A}=\frac{\tan ^{2} A-1}{2 \tan A}=-\frac{1}{\tan 2 A}=\tan B \Rightarrow \tan 2 A=-\frac{1}{\tan B} \\
\Rightarrow 2 A=90^{\circ}+B, \text{ so } \sin 2 A=\cos B=2 \cos ^{2} \frac{B}{2}-1=\frac{2 \sqrt{6}-3}{3} .
\end{array}
$$