Toán Lớp 11: 2sinx .cosx +√3 -2cosx -√3 sinx=0
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Toán Lớp 11: 2sinx .cosx +√3 -2cosx -√3 sinx=0
Comments ( 1 )
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = – \dfrac{\pi }{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
2\sin x.\cos x + \sqrt 3 – 2\cos x – \sqrt 3 \sin x = 0\\
\Leftrightarrow \left( {2\sin x.\cos x – 2\cos x} \right) + \left( { – \sqrt 3 \sin x + \sqrt 3 } \right) = 0\\
\Leftrightarrow 2\cos x.\left( {\sin x – 1} \right) – \sqrt 3 \left( {\sin x – 1} \right) = 0\\
\Leftrightarrow \left( {\sin x – 1} \right)\left( {2\cos x – \sqrt 3 } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x – 1 = 0\\
2\cos x – \sqrt 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\cos x = \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \sin \dfrac{\pi }{2}\\
\cos x = \cos \dfrac{\pi }{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = – \dfrac{\pi }{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)