Toán Lớp 11: giải giúp e với ạ
cos^2x=3sin2x+3
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Toán Lớp 11: giải giúp e với ạ
cos^2x=3sin2x+3
Comments ( 1 )
x = \dfrac{1}{2}\arccos \dfrac{5}{{\sqrt {37} }} – \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt {37} }} + k\pi \\
x = – \dfrac{1}{2}\arccos \dfrac{5}{{\sqrt {37} }} – \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt {37} }} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
{\cos ^2}x = 3\sin 2x + 3\\
\Leftrightarrow 2{\cos ^2}x = 6\sin 2x + 6\\
\Leftrightarrow 2{\cos ^2}x – 1 = 6\sin 2x + 5\\
\Leftrightarrow \cos 2x = 6\sin 2x + 5\\
\Leftrightarrow \cos 2x – 6\sin 2x = 5\\
\Leftrightarrow \dfrac{1}{{\sqrt {37} }}.\cos 2x – \dfrac{6}{{\sqrt {37} }}.\sin 2x = \dfrac{5}{{\sqrt {37} }}\\
\Leftrightarrow \cos 2x.\cos \alpha – \sin 2x.\sin \alpha = \dfrac{5}{{\sqrt {37} }}\\
\left( {\cos \alpha = \dfrac{1}{{\sqrt {37} }};\sin \alpha = \dfrac{6}{{\sqrt {37} }}} \right)\\
\Leftrightarrow \cos \left( {2x + \alpha } \right) = \dfrac{5}{{\sqrt {37} }}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \alpha = \arccos \dfrac{5}{{\sqrt {37} }} + k2\pi \\
2x + \alpha = – \arccos \dfrac{5}{{\sqrt {57} }} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \arccos \dfrac{5}{{\sqrt {37} }} – \alpha + k2\pi \\
2x = – \arccos \dfrac{5}{{\sqrt {37} }} – \alpha + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\arccos \dfrac{5}{{\sqrt {37} }} – \dfrac{1}{2}\alpha + k\pi \\
x = – \dfrac{1}{2}\arccos \dfrac{5}{{\sqrt {37} }} – \dfrac{1}{2}\alpha + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\arccos \dfrac{5}{{\sqrt {37} }} – \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt {37} }} + k\pi \\
x = – \dfrac{1}{2}\arccos \dfrac{5}{{\sqrt {37} }} – \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt {37} }} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)