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$\begin{array}{l}
t = \sin x + \cos x = \sqrt 2 \sin (x + \frac{\pi }{4}),\,\,(\left| t \right| \le \sqrt 2 ) \Rightarrow \sin x\cos x = \frac{{{t^2} - 1}}{2}\\
(*) \Leftrightarrow - 1 + t(1 - \frac{{{t^2} - 1}}{2}) = \frac{3}{2}({t^2} - 1)\\
\Leftrightarrow (t - 1)({t^2} + 4t + 1) = 0\\
\Leftrightarrow {t_1} = 1,\,{t_2} = - 2 + \sqrt 3 ,\,{t_3} = - 2 - \sqrt 3
\end{array}$
Với $t = 1 \Rightarrow \sin (x + \frac{\pi }{4}) = 1 \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \frac{\pi }{2} + k2\pi
\end{array} \right.\,(k \in Z)$
Với $t = \sqrt 3 - 2 \Rightarrow \sin (x + \frac{\pi }{4}) = \frac{{\sqrt 3 - 2}}{{\sqrt 2 }} = \sin \alpha $
$\Leftrightarrow \left[ \begin{array}{l}
x = \alpha - \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} - \alpha + k2\pi
\end{array} \right.\,(\sin \alpha = \frac{{\sqrt 3 - 2}}{{\sqrt 2 }},\,k \in Z)$
Với $ t = - 2 - \sqrt 3 $ (loại)
Vậy nghiệm cần tìm là: $\left[ \begin{array}{l}
x = k2\pi \\
x = \frac{\pi }{2} + k2\pi \\
x = \alpha - \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} - \alpha + k2\pi
\end{array} \right.\,(\sin \alpha = \frac{{\sqrt 3 - 2}}{{\sqrt 2 }},\,k \in Z)$
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Trả lời 18-07-14 08:56 AM
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