Toán Lớp 11: giải pt lượng giác
cos^2(x-30°)=3/4
sin(2x+pi/3)=2sinxcosx
2cos^2-1=2sinxcosx
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Toán Lớp 11: giải pt lượng giác
cos^2(x-30°)=3/4
sin(2x+pi/3)=2sinxcosx
2cos^2-1=2sinxcosx
Comments ( 1 )
1,\\
\left[ \begin{array}{l}
x = 60^\circ + k.360^\circ \\
x = k.180^\circ \\
x = – 120^\circ + k.360^\circ
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
1,\\
{\cos ^2}\left( {x – 30^\circ } \right) = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x – 30^\circ } \right) = \dfrac{{\sqrt 3 }}{2}\\
\cos \left( {x – 30^\circ } \right) = – \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x – 30^\circ = 30^\circ + k.360^\circ \\
x – 30^\circ = – 30^\circ + k.360^\circ \\
x – 30^\circ = 150^\circ + k.360^\circ \\
x – 30^\circ = – 150^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 60^\circ + k.360^\circ \\
x = k.360^\circ \\
x = 180^\circ + k.360^\circ \\
x = – 120^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 60^\circ + k.360^\circ \\
x = k.180^\circ \\
x = – 120^\circ + k.360^\circ
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sin \left( {2x + \dfrac{\pi }{3}} \right) = 2\sin x.\cos x\\
\Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \sin 2x\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = 2x + k2\pi \\
2x + \dfrac{\pi }{3} = \pi – 2x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
0x = – \dfrac{\pi }{3} + k2\pi \,\,\,\,\left( L \right)\\
4x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\\
3,\\
2{\cos ^2}x – 1 = 2\sin x.\cos x\\
\Leftrightarrow \cos 2x = \sin 2x\\
\Leftrightarrow \cos 2x = \cos \left( {\dfrac{\pi }{2} – 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} – 2x + k2\pi \\
2x = 2x – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k2\pi \\
0x = – \dfrac{\pi }{2} + k2\pi \,\,\,\,\,\,\left( L \right)
\end{array} \right.\\
\Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)