Toán Lớp 11: Giúp mình với ạ!!
cosx + √3 sinx+2cosx(2x+π/3)=0
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Toán Lớp 11: Giúp mình với ạ!!
cosx + √3 sinx+2cosx(2x+π/3)=0
Comments ( 1 )
\cos x + \sqrt 3 \sin x + 2\cos \left( {2x + \dfrac{\pi }{3}} \right) = 0\\
\Leftrightarrow 2\left( {\dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x} \right) + 2\cos \left( {2x + \dfrac{\pi }{3}} \right) = 0\\
\Leftrightarrow 2\cos \left( {x – \dfrac{\pi }{3}} \right) + 2\cos \left( {2x + \dfrac{\pi }{3}} \right) = 0\\
\Leftrightarrow \cos \left( {x – \dfrac{\pi }{3}} \right) = – \cos \left( {2x + \dfrac{\pi }{3}} \right) \\
\Leftrightarrow \cos \left( {x – \dfrac{\pi }{3}} \right) = \cos \left( {\pi – 2x – \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \cos \left( {x – \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3} – 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} – 2x + k2\pi \\
x – \dfrac{\pi }{3} = 2x – \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \pi + k2\pi \\
– x = \dfrac{{ – \pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{3} – k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$