Toán Lớp 11: a) sin3x – cos2x =0
b) sin3x = -1/2
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Toán Lớp 11: a) sin3x – cos2x =0
b) sin3x = -1/2
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + \dfrac{{k2\pi }}{5}\\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = – \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{7\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
a,\\
\sin 3x – \cos 2x = 0\\
\Leftrightarrow \cos 2x = \sin 3x\\
\Leftrightarrow \cos 2x = \cos \left( {\dfrac{\pi }{2} – 3x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} – 3x + k2\pi \\
2x = 3x – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
5x = \dfrac{\pi }{2} + k2\pi \\
– x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + \dfrac{{k2\pi }}{5}\\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\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{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{7\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)