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$ \begin{array}{l}
( * ) \Leftrightarrow \sqrt 2 {\sin ^3}\left( {x + {\textstyle{\pi \over 4}}} \right) = 2{\mathop{\rm s}\nolimits} {\rm{inx}}\\
\Leftrightarrow 2\sqrt 2 {\sin ^3}\left( {x + {\textstyle{\pi \over 4}}} \right) = 4{\mathop{\rm s}\nolimits} {\rm{inx}}\\
\Leftrightarrow {\left[ {\sqrt {\rm{2}} \sin \left( {x + {\textstyle{\pi \over 4}}} \right)} \right]^3} = 4\sin x\\
\Leftrightarrow {\left( {{\mathop{\rm s}\nolimits} {\rm{inx + cosx}}} \right)^3} = 4\sin x\,\,\,\left( {**} \right)
\end{array} $
Nếu $cosx = 0$ thì: $ (**) \Leftrightarrow \left\{ \begin{array}{l}
c{\rm{osx = 0}}\\
{\sin ^3}x = 4\sin x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\mathop{\rm s}\nolimits} {\rm{inx = 1}} \vee {\rm{sinx = }} - {\rm{1}}\\
{\sin ^3}x = 4\sin x
\end{array} \right. \Leftrightarrow x \in \emptyset $
Vậy cosx khác 0, ta chia 2 vế của (**) cho $\cos^3x \neq 0$ ta có:
$ \begin{array}{l}
\left( {**} \right) \Leftrightarrow {\left( {{\mathop{\rm t}\nolimits} {\rm{anx + 1}}} \right)^3} = 4\tan x\left( {1 + {{\tan }^2}x} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {\tan ^3}x + 3{\tan ^2}x + 3\tan x + 1 = 4{\tan ^3}x + 4\tan x\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 3{\tan ^3}x - 3{\tan ^2}x + {\mathop{\rm t}\nolimits} {\rm{anx}} - {\rm{1 = 0}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left( {{\rm{tanx}} - {\rm{1}}} \right)\left( {3{{\tan }^2}x + 1} \right) = 0\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {\mathop{\rm t}\nolimits} {\rm{anx = 1}} \Leftrightarrow {\rm{x = }}{\textstyle{\pi \over 4}} + k\pi \,\,,\,\,\,\,\,k \in Z
\end{array} $
Vậy nghiệm cần tìm: $ {\rm{x = }}{\textstyle{\pi \over 4}} + k\pi \,\,,\,\,\,\,\,k \in Z $
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