Toán Lớp 11: $\text{Giải PT: sin x + sin 2x + sin 3x = 0}$
Mn giúp mình với OwO
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Toán Lớp 11: $\text{Giải PT: sin x + sin 2x + sin 3x = 0}$
Mn giúp mình với OwO
Comments ( 1 )
x = \dfrac{\pi }{2} = k\pi \\
x = \dfrac{{k2\pi }}{3}\\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
\sin x + \sin y = 2\sin \dfrac{{x + y}}{2}.\cos \dfrac{{x – y}}{2}\\
\sin 2x = 2\sin x.\cos x\\
\sin x + \sin 2x + \sin 3x = 0\\
\Leftrightarrow \left( {\sin 3x + \sin x} \right) + \sin 2x = 0\\
\Leftrightarrow 2\sin \dfrac{{3x + x}}{2}.\cos \dfrac{{3x – x}}{2} + 2\sin x.\cos x = 0\\
\Leftrightarrow 2\sin 2x.\cos x + 2\sin x.\cos x = 0\\
\Leftrightarrow 2\cos x.\left( {\sin 2x + \sin x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin 2x + \sin x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin 2x = – \sin x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin 2x = \sin \left( { – x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
2x = – x + k2\pi \\
2x = \pi – \left( { – x} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
3x = k2\pi \\
x = \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} = k\pi \\
x = \dfrac{{k2\pi }}{3}\\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)