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Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + {\cos ^2}2x = \dfrac{1}{2}\sin 6x + 1\\
\Leftrightarrow {\sin ^2}x + \left( {1 - {{\sin }^2}2x} \right) = \dfrac{1}{2}\sin 6x + 1\\
\Leftrightarrow {\sin ^2}x - {\sin ^2}2x + 1 = \dfrac{1}{2}.\sin 6x + 1\\
\Leftrightarrow \left( {\sin x - \sin 2x} \right).\left( {\sin x + \sin 2x} \right) = \dfrac{1}{2}\sin 6x\\
\Leftrightarrow 2.\cos \dfrac{{x + 2x}}{2}.\sin \dfrac{{x - 2x}}{2}.2.sin\dfrac{{x + 2x}}{2}.cos\dfrac{{x - 2x}}{2} = \dfrac{1}{2}\sin 6x\\
\Leftrightarrow 4.\cos \dfrac{{3x}}{2}.\sin \dfrac{{ - x}}{2}.sin\dfrac{{3x}}{2}.cos\dfrac{{ - x}}{2} = \dfrac{1}{2}.\sin 6x\\
\Leftrightarrow - 4.sin\dfrac{x}{2}.\cos \dfrac{x}{2}.\sin \dfrac{{3x}}{2}.\cos \dfrac{{3x}}{2} = \dfrac{1}{2}\sin 6x\\
\Leftrightarrow - \left( {2\sin \dfrac{x}{2}.\cos \dfrac{x}{2}} \right).\left( {2\sin \dfrac{{3x}}{2}.\cos \dfrac{{3x}}{2}} \right) = \dfrac{1}{2}.2.\sin 3x.\cos 3x\\
\Leftrightarrow - \sin x.\sin 3x = \sin 3x.\cos 3x\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 3x = 0\\
\cos 3x = - \sin x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = k\pi \\
\cos 3x = \sin \left( { - x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
\cos 3x = \cos \left( {\dfrac{\pi }{2} + x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
3x = \dfrac{\pi }{2} + x + k2\pi \\
3x = - \dfrac{\pi }{2} - x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{4} + k\pi \\
x = - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}
\end{array} \right.
\end{array}\)
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