sin(pi+pi/6)=cos(pi/3-2x) ai giúp em với ạ! 19/07/2021 Bởi Katherine sin(pi+pi/6)=cos(pi/3-2x) ai giúp em với ạ!
Đáp án: $\begin{array}{l}\sin \left( {x + \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{3} – 2x} \right)\\ \Rightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{\pi }{2} – \dfrac{\pi }{3} + 2x} \right)\\ \Rightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{6} = \dfrac{\pi }{2} – \dfrac{\pi }{3} + 2x + k2\pi \\x + \dfrac{\pi }{6} = \pi – \left( {\dfrac{\pi }{2} – \dfrac{\pi }{3} + 2x} \right) + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = – k2\pi \\3x = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}\end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
\sin \left( {x + \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{3} – 2x} \right)\\
\Rightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{\pi }{2} – \dfrac{\pi }{3} + 2x} \right)\\
\Rightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{6} = \dfrac{\pi }{2} – \dfrac{\pi }{3} + 2x + k2\pi \\
x + \dfrac{\pi }{6} = \pi – \left( {\dfrac{\pi }{2} – \dfrac{\pi }{3} + 2x} \right) + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – k2\pi \\
3x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.
\end{array}$