Toán Lớp 11: Giải phương trình
cos2x + căn 3 . sin2x = – căn 3
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Toán Lớp 11: Giải phương trình
cos2x + căn 3 . sin2x = – căn 3
Comments ( 1 )
x = \dfrac{{7\pi }}{{12}} + k\pi \\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
\cos 2x + \sqrt 3 \sin 2x = – \sqrt 3 \\
\Leftrightarrow \dfrac{1}{2}\cos 2x + \dfrac{{\sqrt 3 }}{2}\sin 2x = – \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \cos 2x.\cos \dfrac{\pi }{3} + \sin 2x.\sin \dfrac{\pi }{3} = – \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \cos \left( {2x – \dfrac{\pi }{3}} \right) = \cos \dfrac{{5\pi }}{6}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + k2\pi \\
2x – \dfrac{\pi }{3} = – \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{7\pi }}{6} + k2\pi \\
2x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{12}} + k\pi \\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)