In right triangle $ABC$ with $\angle B = 90^\circ$, we have $2\sin A = 3\cos A$.  What is $\tan A$?
The triangle is shown below:

[asy]
pair A,B,C;
A = (0,0);
B = (10,0);
C = (10,15);
draw(A--B--C--A);
draw(rightanglemark(C,B,A,26));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
[/asy]

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 $\sin A = \frac32\cos A$, so $\tan A = \frac{\sin A}{\cos A} = \frac{(3/2)\cos A}{\cos A } =\boxed{\frac32}$.