Đáp án:
$\begin{array}{l}
Y = \sqrt {2017 - 2017{{(c{\rm{osx)}}}^2}.{{({\mathop{\rm s}\nolimits} {\rm{inx}})}^2}} \\
= \sqrt {2017} .\sqrt {1 - \frac{1}{4}{{(\sin 2x)}^2}} (do\sin 2x = 2\sin x.c{\rm{os}}x)\\
do\, - 1 \le \sin 2x \le 1\\
nen\,0 \le {(\sin 2x)^2} \le 1\\
\to \sqrt {2017} .\sqrt {1 - \frac{1}{4}} \le y \le \sqrt {2017} \\
\to \sqrt {2017} .\sqrt {\frac{3}{4}} \le y \le \sqrt {2017} \\
\to \frac{{\sqrt {6051} }}{2} \le y \le \sqrt {2017} \\
vay\,\min {\rm{ = }}\frac{{\sqrt {6051} }}{2}\,va\,m{\rm{ax}} = \sqrt {2017} \\
\end{array}$
