Toán Lớp 11: giải pt: 2sinmũ2X + sin2x=0 nhanh giúp tôi với ạ
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Toán Lớp 11: giải pt: 2sinmũ2X + sin2x=0 nhanh giúp tôi với ạ
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x = k\pi \\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
2{\sin ^2}x + \sin 2x = 0\\
\Leftrightarrow 2{\sin ^2}x + 2\sin x.\cos x = 0\\
\Leftrightarrow 2\sin x.\left( {\sin x + \cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x + \cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\dfrac{{\sqrt 2 }}{2}.\sin x + \dfrac{{\sqrt 2 }}{2}.\cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x.\cos \dfrac{\pi }{4} + \cos x.\sin \dfrac{\pi }{4} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x + \dfrac{\pi }{4} = k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)