Explanation

(d)
Given, $x + \frac{1}{x} = 2\cos\alpha$
$\Rightarrow \ \ x^{2} - 2x\cos{\alpha + 1 = 0}$
$\Rightarrow \ \ \ x = \frac{2\cos{\alpha \pm \sqrt{4\cos^{2}\alpha - 4}}}{2}$
$\Rightarrow \ \ x = \cos{\alpha + i\sin\alpha}$
Now, $x^{n} = \left( \cos{\alpha + i\sin\alpha} \right)^{n} = \cos{n\alpha + i\sin{n\alpha}}$
And $\frac{1}{x^{n}} = \left( \cos{\alpha - i\sin\alpha} \right)^{n} = \cos{n\alpha - i\sin{n\alpha}}$
$\therefore\ \ \ x^{n} + \frac{1}{x^{n}} = \cos{n\alpha + i\sin{n\alpha + \cos{n\alpha - i\sin{n\alpha}}}}$
$= 2\cos{n\alpha}$