Tìm TXĐ của y=sinx/cos√x^2-2x

$$\eqalign{ & y = {{\sin x} \over {\cos \sqrt {{x^2} - 2x} }} \cr & DKXD:\,\,\left\{ \matrix{ \cos \sqrt {{x^2} - 2x} \ne 0 \hfill \cr {x^2} - 2x \ge 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ \sqrt {{x^2} - 2x} \ne {\pi \over 2} + k\pi \,\,\left( {k \in Z} \right)\,\,\left( 1 \right) \hfill \cr \left[ \matrix{ x \ge 2 \hfill \cr x \le 0 \hfill \cr} \right. \hfill \cr} \right. \cr & Xet\,\,{\pi \over 2} + k\pi < 0\,\,\left( {k \in Z} \right) \cr & \Leftrightarrow {1 \over 2} + k < 0 \Leftrightarrow k < - {1 \over 2} \Rightarrow \left( 1 \right)\,\,luon\,\,dung \cr & Xet\,\,{\pi \over 2} + k\pi \ge 0 \Leftrightarrow k \ge - {1 \over 2} \cr & \left( 1 \right) \Leftrightarrow {x^2} - 2x \ne {\left( {{\pi \over 2} + k\pi } \right)^2} = {a^2} \cr & \Leftrightarrow {x^2} - 2x - {a^2} \ne 0 \cr & \Delta ' = 1 + {a^2} > 0\,\,\forall a \Rightarrow x \ne 1 \pm \sqrt {1 + {a^2}} \,\,voi\,\,a = {\pi \over 2} + k\pi \,\,\left( {k \in Z} \right) \cr} $$