Toán Lớp 11: tan2x -sin²2x= sin2x ( giải , và giải thích dùm tôi nhá mn)
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Toán Lớp 11: tan2x -sin²2x= sin2x ( giải , và giải thích dùm tôi nhá mn)
Comments ( 1 )
\tan 2x – {\sin ^2}2x = \sin 2x\\
\Leftrightarrow \dfrac{{\sin 2x}}{{\cos 2x}} – {\sin ^2}2x = \sin 2x\\
\Leftrightarrow \dfrac{{\sin 2x}}{{\cos 2x}} – {\sin ^2}2x – \sin 2x = 0\\
\Leftrightarrow \sin 2x\left( {\dfrac{1}{{\cos 2x}} – \sin 2x – 1} \right) = 0\\
\Leftrightarrow \sin 2x\left( {\dfrac{{1 – \sin 2x\cos 2x – \cos 2x}}{{\cos 2x}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
1 – \sin 2x\cos 2x – \cos 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
1 – \sin 2x\cos 2x – \left( {1 – 2{{\sin }^2}x} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
2{\sin ^2}x – \sin 2x\cos 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
2{\sin ^2}x = \sin 2x\cos 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\dfrac{{2{{\sin }^2}x}}{{\sin 2x}} = \cos 2x\left( {\sin 2x \ne 0} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\dfrac{{2{{\sin }^2}x}}{{2\sin x\cos x}} = \cos 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\dfrac{{\sin x}}{{\cos x}} = \cos 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\tan x = \cos 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
t = \dfrac{{1 – {t^2}}}{{1 + {t^2}}}\left( {t = \tan x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
{t^3} + t = 1 – {t^2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
\tan x \approx 0,5436
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \arctan \left( {0,5436} \right) + k\pi
\end{array} \right.
\end{array}$