Step-by-step Solution:

Concept:

If \( \alpha, \beta \) are the roots of an equation \( x^2 - 2x \cos \theta + 1 = 0 \), then the equation having \( \alpha^n \) and \( \beta^n \) is:

\[ x^2 - (\alpha^n + \beta^n) x + \alpha^n \beta^n = 0 \]

If \( \alpha = \cos \theta + i \sin \theta \), then by De Moivre's Theorem, we have:

\[ \alpha^n = \cos n\theta + i \sin n\theta \]

Calculations:

Consider the equation \( x^2 - 2x \cos \theta + 1 = 0 \).

\[ x = \frac{2\cos \theta \pm \sqrt{4 \cos^2 \theta - 4(1)(1)}}{2(1)} \]

\[ x = \cos \theta \pm i \sin \theta \]

Given, \( \alpha, \beta \) are the roots of the equation \( x^2 - 2x \cos \theta + 1 = 0 \), we have:

\[ \alpha = \cos \theta + i \sin \theta \quad \text{and} \quad \beta = \cos \theta - i \sin \theta \]

\[ \alpha^n = (\cos \theta + i \sin \theta)^n \quad \text{and} \quad \beta^n = (\cos \theta - i \sin \theta)^n \]

\[ \alpha^n = \cos n\theta + i \sin n\theta \quad \text{and} \quad \beta^n = \cos n\theta - i \sin n\theta \]

\[ \alpha^n + \beta^n = 2\cos n\theta \]

and

\[ \alpha^n \beta^n = 1 \]

The required equation is:

\[ x^2 - (\alpha^n + \beta^n) x + \alpha^n \beta^n = 0 \]

\[ x^2 - 2 \cos n\theta \, x + 1 = 0 \]