Toán Lớp 11: Giải phương trình sau: √3sin6x-2sin4x=cos6x
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Toán Lớp 11: Giải phương trình sau: √3sin6x-2sin4x=cos6x
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x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{7\pi }}{{60}} + \dfrac{{k\pi }}{5}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
\sqrt 3 \sin 6x – 2\sin 4x = \cos 6x\\
\Leftrightarrow \sqrt 3 \sin 6x – \cos 6x = 2\sin 4x\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 6x – \dfrac{1}{2}\cos 6x = \sin 4x\\
\Leftrightarrow \sin 6x.\cos \dfrac{\pi }{6} – \cos 6x.\sin \dfrac{\pi }{6} = \sin 4x\\
\Leftrightarrow \sin \left( {6x – \dfrac{\pi }{6}} \right) = \sin 4x\\
\Leftrightarrow \left[ \begin{array}{l}
6x – \dfrac{\pi }{6} = 4x + k2\pi \\
6x – \dfrac{\pi }{6} = \pi – 4x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
6x – 4x = \dfrac{\pi }{6} + k2\pi \\
6x + 4x = \pi + \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
10x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{7\pi }}{{60}} + \dfrac{{k\pi }}{5}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)