Toán Lớp 11: Giải pt 2sin(2x+π/9)+√3=0
Giải pt 2sin 2x – 2cos2x=√2
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Toán Lớp 11: Giải pt 2sin(2x+π/9)+√3=0
Giải pt 2sin 2x – 2cos2x=√2
Comments ( 2 )
a)\\
2\sin \left( {2x + \dfrac{\pi }{9}} \right) + \sqrt 3 = 0\\
\Leftrightarrow \sin \left( {2x + \dfrac{\pi }{9}} \right) = – \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{9} = – \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{\pi }{9} = \pi + \dfrac{\pi }{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = – \dfrac{{4\pi }}{9} + k2\pi \\
2x = \dfrac{{11\pi }}{9} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{{2\pi }}{9} + k\pi \\
x = \dfrac{{11\pi }}{{18}} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,\left[ \begin{array}{l}
x = – \dfrac{{2\pi }}{9} + k\pi \\
x = \dfrac{{11\pi }}{{18}} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
2)2\sin 2x – 2\cos 2x = \sqrt 2 \\
\Leftrightarrow \sin 2x – \cos 2x = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sqrt 2 \sin \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sin \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\\
\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.\left( {k \in Z} \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)\\
Vậy\,\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)
\end{array}$