Đặt \(t = \tan \frac{x}{2}\)
\( \Rightarrow \sin x = \frac{{2t}}{{1 + {t^2}}};\,\,\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}\)
\( \Rightarrow y = k\left( {\frac{{\frac{{2t}}{{1 + {t^2}}}}}{{\frac{{2t + 1 - {t^2} + 2 + 2t}}{{1 + {t^2}}}}}} \right) = k\left( {\frac{{2t}}{{{t^2} + 2t + 3}}} \right)\)
\(y' = k\left( {\frac{{ - 2{t^2} + 6}}{{{{\left( {{t^2} + 2t + 3} \right)}^2}}}} \right)\)
\(y' = 0 \Leftrightarrow \left[ \begin{array}{l}k = 0\,(loai)\\ - 2{t^2} + 6 = 0\end{array} \right. \Rightarrow t = \pm \sqrt 3 \)
\( \Rightarrow \left[ \begin{array}{l}\tan \frac{x}{2} = \sqrt 3 \\\tan \frac{x}{2} = - \sqrt 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{2\pi }}{3}\\x = - \frac{{2\pi }}{3}\end{array} \right.\)
Miền giá trị của y:
\(d = k\left| {\frac{{\sin \left( {\frac{{2\pi }}{3}} \right)}}{{\cos \left( {\frac{{2\pi }}{3}} \right) + \sin \left( {\frac{{2\pi }}{3}} \right) + 2}} - \frac{{\sin \left( { - \frac{{2\pi }}{3}} \right)}}{{\cos \left( { - \frac{{2\pi }}{3}} \right) + \sin \left( { - \frac{{2\pi }}{3}} \right) + 2}}} \right| = k.\sqrt 3 \)
\(d = 3 \Rightarrow k.\sqrt 3 = 3 \Rightarrow k = \sqrt 3 .\)
