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Điều kiện: $ \sin 2x \ne 0\,;\,{\mathop{\rm t}\nolimits} {\rm{anx}} \ne - 1$ (**).
Ta có:
$ \begin{array}{l}
\frac{{c{\rm{os}}2x}}{{1 + {\mathop{\rm t}\nolimits} {\rm{anx}}}} = \frac{{c{\rm{o}}{{\rm{s}}^2}x - {{\sin }^2}x}}{{1 + \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{{\cos x}}}} = \frac{{\cos x\left( {c{\rm{o}}{{\rm{s}}^2}x - {{\sin }^2}x} \right)}}{{\cos x + {\mathop{\rm s}\nolimits} {\rm{inx}}}} = \cos x(\cos x - {\mathop{\rm s}\nolimits} {\rm{inx}})\\
\end{array} $
Do đó:
$\begin{array}{l}
\left( * \right) \Leftrightarrow cotx - 1 = \left( {c{\rm{o}}{{\rm{s}}^2}x - \sin {\rm{x}}\cos x} \right) + {\sin ^2}x - \frac{1}{2}\sin 2x\\
\Leftrightarrow \frac{{\cos x - {\mathop{\rm s}\nolimits} {\rm{inx}}}}{{{\mathop{\rm s}\nolimits} {\rm{inx}}}} = 1 - \sin 2x\\
\Leftrightarrow \cos x - {\mathop{\rm s}\nolimits} {\rm{inx}} = {\mathop{\rm s}\nolimits} {\rm{inx}}{\left( {\cos x - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x - {\mathop{\rm s}\nolimits} {\rm{inx}} = 0\\
1 = {\mathop{\rm s}\nolimits} {\rm{inx}}\left( {\cos x - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{\mathop{\rm t}\nolimits} {\rm{anx}} = 1\\
\frac{1}{{c{\rm{o}}{{\rm{s}}^2}x}} = \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{{\cos x}} - {\tan ^2}x\,\,\,({\rm{do}}\,\,\cos x \ne 0)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + k\pi \,\,(k \in Z\,)\\
2{\tan ^2}x - {\mathop{\rm t}\nolimits} {\rm{anx}} + 1 = 0\,(\,{\rm{VN}})
\end{array} \right.\\
\Leftrightarrow x = \frac{\pi }{4} + k\pi ;k \in Z
\end{array}$
Đối chiếu với điều kiện (**), ta có nghiệm cần tìm: $x = \frac{\pi }{4} + k\pi ;k \in Z$
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