Toán Lớp 11: Giải giúp em câu này hứa cho 5 sao ạ
tan(x+π/6)=-cot(2x-π/3)
Giải từ đầu đến cuối luôn dùm em nha
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: Giải giúp em câu này hứa cho 5 sao ạ
tan(x+π/6)=-cot(2x-π/3)
Giải từ đầu đến cuối luôn dùm em nha
Comments ( 1 )
\left\{ \begin{array}{l}
\cos \left( {x + \dfrac{\pi }{6}} \right) \ne 0\\
\sin \left( {2x – \dfrac{\pi }{3}} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + \dfrac{\pi }{6} \ne \dfrac{\pi }{2} + k\pi \\
2x – \dfrac{\pi }{3} \ne k\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{3} + k\pi \\
2x \ne \dfrac{\pi }{3} + k\pi
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{3} + k\pi \\
x \ne \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
\cot x = – \cot \left( { – x} \right)\\
\cot x = \tan \left( {\dfrac{\pi }{2} – x} \right)\\
\tan \left( {x + \dfrac{\pi }{6}} \right) = – \cot \left( {2x – \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \tan \left( {x + \dfrac{\pi }{6}} \right) = \cot \left( {\dfrac{\pi }{3} – 2x} \right)\\
\Leftrightarrow \tan \left( {x + \dfrac{\pi }{6}} \right) = \tan \left[ {\dfrac{\pi }{2} – \left( {\dfrac{\pi }{3} – 2x} \right)} \right]\\
\Leftrightarrow \tan \left( {x + \dfrac{\pi }{6}} \right) = \tan \left( {2x + \dfrac{\pi }{6}} \right)\\
\Leftrightarrow x + \dfrac{\pi }{6} = 2x + \dfrac{\pi }{6} + k\pi \\
\Leftrightarrow x = k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)