Toán Lớp 11: giải giúp e câu này ạ em cảm ơn
sin^2x+sinx^2x=sin^23x
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Toán Lớp 11: giải giúp e câu này ạ em cảm ơn
sin^2x+sinx^2x=sin^23x
Comments ( 2 )
{\sin ^2}x + {\sin ^2}2x = {\sin ^2}3x\\
\Leftrightarrow \dfrac{{1 – \cos 2x}}{2} + \dfrac{{1 – \cos 4x}}{2} = \dfrac{{1 – \cos 6x}}{2}\\
\Leftrightarrow 1 – \dfrac{{\left( {\cos 2x + \cos 4x} \right)}}{2} = \dfrac{1}{2} – \dfrac{{\cos 6x}}{2}\\
\Leftrightarrow \dfrac{{\cos 2x + \cos 4x}}{2} – \dfrac{{\cos 6x}}{2} – \dfrac{1}{2} = 0\\
\Leftrightarrow \cos 2x + \cos 4x – \cos 6x – 1 = 0\\
\Leftrightarrow 2\cos 3x\cos x – \left( {2{{\cos }^2}3x – 1} \right) – 1 = 0\\
\Leftrightarrow 2\cos 3x\left( {\cos x – \cos 3x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 3x = 0\\
\cos 3x = \cos x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} + k\pi \\
3x = x + k2\pi \\
3x = – x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\\
x = k\pi \\
x = \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$