Toán Lớp 11: Tìm x thuộc (-pi,pi) sao cho sin(x-pi/3)+2cos(x+pi/6)=0
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Toán Lớp 11: Tìm x thuộc (-pi,pi) sao cho sin(x-pi/3)+2cos(x+pi/6)=0
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x = – \dfrac{{2\pi }}{3}\\
x = \dfrac{\pi }{3}
\end{array} \right.\]
\sin \left( {x – \dfrac{\pi }{3}} \right) + 2\cos \left( {x + \dfrac{\pi }{6}} \right) = 0\\
\Leftrightarrow \sin \left( {x – \dfrac{\pi }{3}} \right) + 2.\sin \left[ {\dfrac{\pi }{2} – \left( {x + \dfrac{\pi }{6}} \right)} \right] = 0\\
\Leftrightarrow \sin \left( {x – \dfrac{\pi }{3}} \right) + 2.\sin \left( {\dfrac{\pi }{3} – x} \right) = 0\\
\Leftrightarrow \sin \left( {x – \dfrac{\pi }{3}} \right) – 2\sin \left( {x – \dfrac{\pi }{3}} \right) = 0\\
\Leftrightarrow – \sin \left( {x – \dfrac{\pi }{3}} \right) = 0\\
\Leftrightarrow \sin \left( {x – \dfrac{\pi }{3}} \right) = 0\\
\Leftrightarrow x – \dfrac{\pi }{3} = k\pi \\
\Leftrightarrow x = \dfrac{\pi }{3} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
x \in \left( { – \pi ;\pi } \right) \Rightarrow – \pi < x < \pi \\
\Leftrightarrow – \pi < \dfrac{\pi }{3} + k\pi < \pi \\
\Leftrightarrow – \dfrac{{4\pi }}{3} < k\pi < \dfrac{{2\pi }}{3}\\
\Leftrightarrow – \dfrac{4}{3} < k < \dfrac{2}{3}\\
k \in Z \Rightarrow \left[ \begin{array}{l}
k = – 1\\
k = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = – \dfrac{{2\pi }}{3}\\
x = \dfrac{\pi }{3}
\end{array} \right.
\end{array}\)