$y=\dfrac{x^2.cos \alpha - 2x + cos \alpha}{x^2-2x.cos \alpha + 1}$
$\Rightarrow x^2y-2xy.cos \alpha + y = x^2.cos \alpha - 2x + cos \alpha$
$\Leftrightarrow x^2(y-cos \alpha)-2x(y.cos \alpha - 1) +y-cos \alpha=0$
$\Delta'=(y.cos \alpha - 1)^2-(y-cos \alpha)^2 \ge 0$
$\Leftrightarrow y^2.cos^2 \alpha-2y.cos \alpha+1-y^2+2y.cos \alpha-cos^2 \alpha \ge 0$
$\Leftrightarrow y^2(cos^2 \alpha -1)-cos^2 \alpha+1 \ge 0$
$\Leftrightarrow y^2 \le \dfrac{cos^2 \alpha -1}{cos^2 \alpha-1} = 1$
$\Leftrightarrow -1 \le y \le 1$
$min(y)=-1 \Leftrightarrow x=1$ và $\alpha \ne 0^o$
$max(y)=1 \Leftrightarrow x=-1$