$\alpha = \sin {36}^o$ is a root of which of the following equation 

To determine the equation for which α = sin 36° is a root, we proceed as follows:

1. Let α = sin 36° = x. Thus, we represent sin 36° with the variable x.
2. We know that x satisfies the relation: x = (10 - 2√5) / 4.
3. Rewriting this, we multiply through by 4: 4x = 10 - 2√5.
4. Square both sides: (4x)^2 = (10 - 2√5)^2, which simplifies to: 
   16x² = 100 - 40√5 + 20.
5. Combine like terms: 16x² = 10 - 2√5.
6. Introducing this into a polynomial form: (8x² - 5)^2 = 5.
7. Expand the expression: 64x⁴ - 80x² + 25 = 5.
8. Simplify further to get: 64x⁴ - 80x² + 20 = 0.
9. Divide through by 4 for simplicity: 16x⁴ - 20x² + 5 = 0.

Hence, the equation for which α = sin 36° is a root is:

16x⁴ - 20x² + 5 = 0

This equation clearly demonstrates that sin 36° is a valid root. It is derived from the fundamental properties of trigonometric values for a 36-degree angle and follows a systematic approach to simplifying the relationship between sin 36° and polynomial equations.