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HPT\( \Leftrightarrow \left\{ \begin{array}{l}
2{\cos ^2}x = 1 + 2\cos y + c{\rm{o}}{{\rm{s}}^2}y\\
2{\sin ^2}x = {\sin ^2}y
\end{array} \right.\)
Cộng
$2$ vế \(\Rightarrow 2 = 1 + \cos y + c{\rm{o}}{{\rm{s}}^2}y + {\sin
^2}y \Rightarrow \cos y = 0 \Leftrightarrow y = \frac{\pi }{2} + k\pi \)
Thế vào hệ đã cho ta được \(\left\{ \begin{array}{l}
\cos x = \frac{1}{{\sqrt 2 }}\\
{\mathop{\rm s}\nolimits} {\rm{inx}} = \frac{{{{\left( { - 1} \right)}^k}}}{{\sqrt 2 }}
\end{array} \right.\)
Nếu $k$ chẵn \(\left( {k = 2l} \right)\) thì ta có: \(\left\{ \begin{array}{l}
\cos x = \frac{{\sqrt 2 }}{2}\\
{\mathop{\rm s}\nolimits} {\rm{inx}} = \frac{{\sqrt 2 }}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{\pi }{4} + 2m\pi \\
y = \frac{\pi }{2} + 2l\pi
\end{array} \right.\left( {m,l \in Z} \right)\)
Nếu $k$ lẻ \(\left( {k = 2l + 1} \right)\) thì ta có: \(\left\{ \begin{array}{l}
\cos x = \frac{{\sqrt 2 }}{2}\\
{\mathop{\rm s}\nolimits} {\rm{inx}} = \frac{{ - \sqrt 2 }}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \frac{\pi }{4} + 2m\pi \\
y = \frac{\pi }{2} + \left( {2l + 1} \right)\pi
\end{array} \right.\left( {m,l \in Z} \right)\)
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