Ta có:
$x^2-2x\sin(xy)+1=0$
$\Leftrightarrow x^2-2x\sin(xy)+\sin^2(xy)+\cos^2(xy)=0$
$\Leftrightarrow (x-\sin(xy))^2+\cos^2(xy)=0$
$\Leftrightarrow \left\{\begin{array}{l}x=\sin(xy)\\cos(xy)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=1\\\sin y=1\\\cos y=0\end{array}\right.\\\left\{\begin{array}{l}x=-1\\\sin(-y)=-1\\\cos(-y)=0\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=1\\y=\dfrac{\pi}{2}+k2\pi\end{array}\right., k\in\mathbb{Z}\\\left\{\begin{array}{l}x=-1\\y=\dfrac{\pi}{2}+k2\pi\end{array}\right., k\in\mathbb{Z}\end{array}\right.$