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

[asy]
pair A,B,C;
A = (0,0);
B = (5,0);
C = (5,10);
draw(A--B--C--A);
draw(rightanglemark(C,B,A,16));
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 $\sin A = 2\cos A$ gives us $\frac{BC}{AC} = 2\cdot\frac{AB}{AC}$.  Multiplying both sides by $AC$ gives $BC = 2AB$.

The Pythagorean Theorem gives us $AB^2 + BC^2 = AC^2$.  Substituting $BC = 2AB$ gives \[AB^2 + (2AB)^2 = AC^2.\]Simplifying the left side gives $5AB^2 = AC^2$, so $\frac{AB^2}{AC^2} = \frac{1}{5}$, which means \[\cos A = \frac{AB}{AC} = \sqrt{\frac{1}{5}} = \frac{\sqrt{1}}{\sqrt{5}} = \frac{1}{\sqrt{5}} = \boxed{\frac{\sqrt{5}}{5}}.\]