QJEEM22Ma7cef: JEE Main 2022 Mathematics - Class 11 JEEM22Ma7cef Solution: is a root of which of the following equation?

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.