Given that $\alpha = \sin 36^\circ$, we need to determine which equation it is a root of.

We start with the known relationship for $\cos 72^\circ$:

$ \cos 72^\circ = \frac{\sqrt{5}-1}{4} $

Using the double-angle formula for cosine:

$ \cos 72^\circ = 1 - 2 \sin^2 36^\circ $

Substitute $\alpha$ for $\sin 36^\circ$:

$ 1 - 2\alpha^2 = \frac{\sqrt{5}-1}{4} $

Multiply both sides by 4:

$ 4 - 8\alpha^2 = \sqrt{5} - 1 $

Add 1 to both sides:

$ 5 - 8\alpha^2 = \sqrt{5} $

Square both sides to eliminate the radical:

$ (5 - 8\alpha^2)^2 = 5 $

Expand the left side:

$ 25 + 64\alpha^4 - 80\alpha^2 = 5 $

Simplify by subtracting 5 from both sides:

$ 64\alpha^4 - 80\alpha^2 + 20 = 0 $

Divide the entire equation by 4:

$ 16\alpha^4 - 20\alpha^2 + 5 = 0 $

Thus, the equation $16\alpha^4 - 20\alpha^2 + 5 = 0$ is the one for which $\alpha = \sin 36^\circ$ is a root.