Toán Lớp 11: Giải phương trình :
a) sin (3x+pi/2)=1/2
b) sin (x-pi/5)=-1
c) sin (2x-pi/4)=√2/2
Mn giải giúp em vs ạ em đang cần gấp ạ
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Toán Lớp 11: Giải phương trình :
a) sin (3x+pi/2)=1/2
b) sin (x-pi/5)=-1
c) sin (2x-pi/4)=√2/2
Mn giải giúp em vs ạ em đang cần gấp ạ
Comments ( 1 )
a)\sin \left( {3x + \dfrac{\pi }{2}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {3x + \dfrac{\pi }{2}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{2} = \dfrac{\pi }{6} + k2\pi \\
3x + \dfrac{\pi }{2} = \pi – \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ – \pi }}{9} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{ – \pi }}{9} + \dfrac{{k2\pi }}{3};x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}\\
b)\sin \left( {x – \dfrac{\pi }{5}} \right) = – 1\\
\Leftrightarrow x – \dfrac{\pi }{5} = \dfrac{{ – \pi }}{2} + k2\pi \\
\Leftrightarrow x = \dfrac{{ – 3\pi }}{{10}} + k2\pi \\
Vậy\,x = \dfrac{{ – 3\pi }}{{10}} + k2\pi \\
c)\sin \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
2x – \dfrac{\pi }{4} = \pi – \dfrac{\pi }{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
Vậy\,x = \dfrac{\pi }{4} + k\pi ;x = \dfrac{\pi }{2} + k\pi
\end{array}$