Toán Lớp 11: giúp e đi mà
sin4x – cos6x = √3( sin6x + cos4x)
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 11: giúp e đi mà
sin4x – cos6x = √3( sin6x + cos4x)
Comments ( 1 )
\sin 4x – \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 4x} \right)\\
\Leftrightarrow \dfrac{1}{2}\sin 4x – \dfrac{{\sqrt 3 }}{2}\cos 4x = \dfrac{{\sqrt 3 }}{2}\sin 6x + \dfrac{1}{2}\cos 6x\\
\Leftrightarrow \sin \left( {4x – \dfrac{\pi }{3}} \right) = \sin \left( {6x + \dfrac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
4x – \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
4x – \dfrac{\pi }{3} = \pi – 6x – \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = – \dfrac{\pi }{2} – k2\pi \\
10x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{7\pi }}{{60}} + \dfrac{{k\pi }}{5}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$