Toán Lớp 11: HELP !!!!
Giải các phương trình sau
a) cot 2x = tan(x+ $\frac{\pi}{6}$)
b) tan( $\frac{\pi}{3}$ -x) = -tan3x
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: HELP !!!!
Giải các phương trình sau
a) cot 2x = tan(x+ $\frac{\pi}{6}$)
b) tan( $\frac{\pi}{3}$ -x) = -tan3x
Comments ( 2 )
Bạn tham khảo nha
Mình thiếu đk á 🙁
a,\\
x = \dfrac{\pi }{9} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
b,\\
x = – \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\sin 2x \ne 0\\
\cos \left( {x + \dfrac{\pi }{6}} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x \ne k\pi \\
x + \dfrac{\pi }{6} \ne \dfrac{\pi }{2} + k\pi
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{{k\pi }}{2}\\
x \ne \dfrac{\pi }{3} + k\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
\cot 2x = \tan \left( {x + \dfrac{\pi }{6}} \right)\\
\Leftrightarrow \tan \left( {\dfrac{\pi }{2} – 2x} \right) = \tan \left( {x + \dfrac{\pi }{6}} \right)\\
\Leftrightarrow \dfrac{\pi }{2} – 2x = x + \dfrac{\pi }{6} + k\pi \\
\Leftrightarrow – 2x – x = \dfrac{\pi }{6} – \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow – 3x = – \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
b,\\
DKXD:\,\,\left\{ \begin{array}{l}
\cos \left( {\dfrac{\pi }{3} – x} \right) \ne 0\\
\cos 3x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{\pi }{3} – x \ne \dfrac{\pi }{2} + k\pi \\
3x \ne \dfrac{\pi }{2} + k\pi
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne – \dfrac{\pi }{6} + k\pi \\
x \ne \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
\tan \left( {\dfrac{\pi }{3} – x} \right) = – \tan 3x\\
\Leftrightarrow \tan \left( {\dfrac{\pi }{3} – x} \right) = \tan \left( { – 3x} \right)\\
\Leftrightarrow \dfrac{\pi }{3} – x = – 3x + k\pi \\
\Leftrightarrow \dfrac{\pi }{3} – x + 3x = k\pi \\
\Leftrightarrow 2x = – \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow x = – \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)