Toán Lớp 11: b) cos x – sin x = căn(2) * sin 2x
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Toán Lớp 11: b) cos x – sin x = căn(2) * sin 2x
Comments ( 1 )
x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
\cos x – \sin x = \sqrt 2 .\sin 2x\\
\Leftrightarrow \dfrac{{\sqrt 2 }}{2}.\cos x – \dfrac{{\sqrt 2 }}{2}\sin x = \dfrac{{\sqrt 2 }}{2}.\sqrt 2 \sin 2x\\
\Leftrightarrow \cos x.\cos \dfrac{\pi }{4} – \sin x.\sin \dfrac{\pi }{4} = \sin 2x\\
\Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \sin 2x\\
\Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{2} – 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \dfrac{\pi }{2} – 2x + k2\pi \\
x + \dfrac{\pi }{4} = 2x – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{4} + k2\pi \\
– x = – \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)