Toán Lớp 11: $\frac{(sin2x -1)(2cosx – √2 )}{sinx – cosx}$ =0
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Toán Lớp 11: $\frac{(sin2x -1)(2cosx – √2 )}{sinx – cosx}$ =0
Comments ( 2 )
Dkxd:\sin x – \cos x \ne 0\\
\Leftrightarrow \sqrt 2 \sin \left( {x – \dfrac{\pi }{4}} \right) \ne 0\\
\Leftrightarrow x – \dfrac{\pi }{4} \ne k\pi \\
\Leftrightarrow x \ne \dfrac{\pi }{4} + k\pi \\
\dfrac{{\left( {\sin 2x – 1} \right)\left( {2\cos x – \sqrt 2 } \right)}}{{\sin x – \cos x}} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x – 1 = 0\\
2\cos x – \sqrt 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 1\\
\cos x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{4} + k2\pi \left( {ktm} \right)\\
x = – \dfrac{\pi }{4} + k2\pi \left( {tm} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \left( {ktm} \right)\\
x = – \dfrac{\pi }{4} + k2\pi
\end{array} \right.\\
Vậy\,x = – \dfrac{\pi }{4} + k2\pi
\end{array}$