Toán Lớp 11: Tìm nghiệm của phương trình sin3x=2sin(x+π/3).sin(π/6-x)
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Toán Lớp 11: Tìm nghiệm của phương trình sin3x=2sin(x+π/3).sin(π/6-x)
Mọi người chỉ mik cách giải với
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x = \dfrac{\pi }{{15}} + \dfrac{{k2\pi }}{5}\\
x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\)
2\sin x.\sin y = – \left[ {\cos \left( {x + y} \right) – \cos \left( {x – y} \right)} \right]\\
\sin 3x = 2\sin \left( {x + \dfrac{\pi }{3}} \right).\sin \left( {\dfrac{\pi }{6} – x} \right)\\
\Leftrightarrow \sin 3x = – \left[ {\cos \left( {x + \dfrac{\pi }{3} + \dfrac{\pi }{6} – x} \right) – \cos \left( {x + \dfrac{\pi }{3} – \dfrac{\pi }{6} + x} \right)} \right]\\
\Leftrightarrow \sin 3x = – \left[ {\cos \dfrac{\pi }{2} – \cos \left( {2x + \dfrac{\pi }{6}} \right)} \right]\\
\Leftrightarrow \sin 3x = – \left[ {0 – \cos \left( {2x + \dfrac{\pi }{6}} \right)} \right]\\
\Leftrightarrow \sin 3x = \cos \left( {2x + \dfrac{\pi }{6}} \right)\\
\Leftrightarrow \cos \left( {2x + \dfrac{\pi }{6}} \right) = \sin 3x\\
\Leftrightarrow \cos \left( {2x + \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{2} – 3x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = \dfrac{\pi }{2} – 3x + k2\pi \\
2x + \dfrac{\pi }{6} = 3x – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 3x = \dfrac{\pi }{2} – \dfrac{\pi }{6} + k2\pi \\
2x – 3x = – \dfrac{\pi }{2} – \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
5x = \dfrac{\pi }{3} + k2\pi \\
– x = – \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{15}} + \dfrac{{k2\pi }}{5}\\
x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)