jmc

The triangle is shown below:



We have $\sin A=\frac{BC}{AC}$ and $\cos A=\frac{AB}{AC}$, so $2\sin A=3\cos B$ gives us $2\cdot \frac{BC}{AC}=3\cdot \frac{AB}{AC}$. Multiplying both sides by $AC$ gives $2BC=3AB$, so $\frac{BC}{AB}=\frac{3}{2}$. Finally, we have $\tan A=\frac{BC}{AB}=\boxed{\frac{3}{2}}$.

We also could have noted that $2\sin A=3\cos A$ gives us $\sin A=\frac{3}{2}\cos A$, so $\tan A=\frac{\sin A}{\cos A}=\frac{(3/2)\cos A}{\cos A}=\boxed{\frac{3}{2}}$.