Toán Lớp 11: Giải các phương trình sau:
a) 2sin3x -1=0
b) 2sin(2x-π/4)-1=0
c) 2cos(2x-π/4)-1=0
d) 2tan(2x-π/4)-1=0
e) √3 cot(x-π/4)-1=0
ai giúp mình với ạ
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 các phương trình sau:
a) 2sin3x -1=0
b) 2sin(2x-π/4)-1=0
c) 2cos(2x-π/4)-1=0
d) 2tan(2x-π/4)-1=0
e) √3 cot(x-π/4)-1=0
ai giúp mình với ạ
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{24}} + k\pi \\
x = \dfrac{{13\pi }}{{24}} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{24}} + k\pi \\
x = – \dfrac{\pi }{{24}} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
x = \dfrac{\pi }{8} + \dfrac{1}{2}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)\\
e,\\
x = \dfrac{{7\pi }}{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
a,\\
2\sin 3x – 1 = 0\\
\Leftrightarrow \sin 3x = \dfrac{1}{2}\\
\Leftrightarrow \sin 3x = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \pi – \dfrac{\pi }{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
2\sin \left( {2x – \dfrac{\pi }{4}} \right) – 1 = 0\\
\Leftrightarrow \sin \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {2x – \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{4} = \dfrac{\pi }{6} + k2\pi \\
2x – \dfrac{\pi }{4} = \pi – \dfrac{\pi }{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{4} = \dfrac{\pi }{6} + k2\pi \\
2x – \dfrac{\pi }{4} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{5\pi }}{{12}} + k2\pi \\
2x = \dfrac{{13\pi }}{{12}} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{24}} + k\pi \\
x = \dfrac{{13\pi }}{{24}} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
2\cos \left( {2x – \dfrac{\pi }{4}} \right) – 1 = 0\\
\Leftrightarrow \cos \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \cos \left( {2x – \dfrac{\pi }{4}} \right) = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{4} = \dfrac{\pi }{3} + k2\pi \\
2x – \dfrac{\pi }{4} = – \dfrac{\pi }{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{7\pi }}{{12}} + k2\pi \\
2x = – \dfrac{\pi }{{12}} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{24}} + k\pi \\
x = – \dfrac{\pi }{{24}} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
DKXD:\,\,\,\cos \left( {2x – \dfrac{\pi }{4}} \right) \ne 0\\
\Leftrightarrow 2x – \dfrac{\pi }{4} \ne \dfrac{\pi }{2} + k\pi \Leftrightarrow x \ne \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
2\tan \left( {2x – \dfrac{\pi }{4}} \right) – 1 = 0\\
\Leftrightarrow \tan \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\\
\Leftrightarrow 2x – \dfrac{\pi }{4} = \arctan \dfrac{1}{2} + k\pi \\
\Leftrightarrow 2x = \dfrac{\pi }{4} + \arctan \dfrac{1}{2} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{1}{2}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)\\
e,\\
DKXD:\,\,\sin \left( {x – \dfrac{\pi }{4}} \right) \ne 0 \Leftrightarrow x – \dfrac{\pi }{4} \ne k\pi \Leftrightarrow x \ne \dfrac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right)\\
\sqrt 3 \cot \left( {x – \dfrac{\pi }{4}} \right) – 1 = 0\\
\Leftrightarrow \cot \left( {x – \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 3 }}\\
\Leftrightarrow x – \dfrac{\pi }{4} = \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow x = \dfrac{{7\pi }}{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)