Slovenia — National Math Olympiad

Let us first consider the case $x > 0$. We have $(x - 1)^2 \ge 0$ or $x^2 + 1 \ge 2x$, and dividing by $x$ we get
$$x + \frac{1}{x} \ge 2 \ge 2 \cos \alpha = x + \frac{1}{x},$$
so $\cos \alpha = 1$ and $x+1/x = 2$. We conclude that $x = 1$. This implies $x^n+1/x^n = 1+1=2 = \cos(n\alpha)$.

If $x < 0$, then a similar reasoning for $(x+1)^2 \ge 0$ shows that $x = -1$ and $\cos \alpha = -1$, whence $x^n + 1/x^n = 2(-1)^n = 2\cos(n\alpha)$.