Toán Lớp 11: cosx.cos2x=1/4
Giúp mik vs ak
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Toán Lớp 11: cosx.cos2x=1/4
Giúp mik vs ak
Comments ( 2 )
\cos 2x\cos x = \dfrac{1}{4}\\
\Leftrightarrow \left( {2{{\cos }^2}x – 1} \right)\cos x = \dfrac{1}{4}\\
\Leftrightarrow 2{\cos ^3}x – \cos x = \dfrac{1}{4}\\
\Leftrightarrow 8{\cos ^3}x – 4\cos x – 1 = 0\\
\Leftrightarrow 8{\cos ^3}x + 4{\cos ^2}x – 4{\cos ^2}x – 2\cos x – 2\cos x – 1 = 0\\
\Leftrightarrow 4{\cos ^2}x\left( {2\cos x + 1} \right) – 2\cos x\left( {2\cos x + 1} \right) – \left( {2\cos x + 1} \right) = 0\\
\Leftrightarrow \left( {2\cos x + 1} \right)\left( {4{{\cos }^2}x – 2\cos x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = – \dfrac{1}{2}(TM)\\
\cos x = \dfrac{{1 + \sqrt 5 }}{4}(L)\\
\cos x = \dfrac{{1 – \sqrt 5 }}{4}(TM)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{{2\pi }}{3} + k2\pi \\
x = \pm \arccos \left( {\dfrac{{1 – \sqrt 5 }}{4}} \right) + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$