Explanation

(b)
The given equation is $x^{2} - 2x\cos{\varnothing + 1 = 0}$
$\therefore x = \frac{2\cos{\varnothing \pm \sqrt{4\cos^{2}{\varnothing - 4}}}}{2} = \cos{\varnothing \pm i\sin\varnothing}$
Let $\alpha = \cos{\varnothing + i\sin\varnothing,\ }$ then $\beta = \cos{\varnothing - i\sin\varnothing}$
$\therefore\ \ \ \alpha^{n} + \beta^{n} = \left( \cos{\varnothing + i\sin\varnothing} \right)^{n} + \left( \cos{\varnothing - i\sin\varnothing} \right)^{n}$
$= 2\cos{n\varnothing}$
And $\alpha^{n}\beta^{n} = \left( \cos{n\varnothing + i\sin{n\varnothing}} \right)(\cos{n\varnothing - i\sin{n\varnothing}})$
$= \cos^{2}{n\varnothing + \sin^{2}{n\varnothing = 1}}$
$\therefore$ Required equation is $x^{2} - 2x\cos{n\varnothing + 1 = 0}$