Toán Lớp 11: Giải phương trình
1. sinx.cos2x + $cos^{2}x$ + $cos^{2}x$ ($tan^{2}x$ – 1) + $2sin^{3}x$ =0
2. cos3x+cos2x-cosx-1=0
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Toán Lớp 11: Giải phương trình
1. sinx.cos2x + $cos^{2}x$ + $cos^{2}x$ ($tan^{2}x$ – 1) + $2sin^{3}x$ =0
2. cos3x+cos2x-cosx-1=0
Comments ( 1 )
\sin x\cos 2x + {\cos ^2}x + {\cos ^2}x\left( {{{\tan }^2}x – 1} \right) + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{1}{2}\left( {\sin 3x – \sin x} \right) + {\cos ^2}x + {\cos ^2}x.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} – {\cos ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{1}{2}\sin 3x – \dfrac{1}{2}\sin x + {\sin ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{1}{2}\left( {3\sin x – 4{{\sin }^3}x} \right) – \dfrac{1}{2}\sin x + {\sin ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{3}{2}\sin x – 2{\sin ^3}x – \dfrac{1}{2}\sin x + {\sin ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow {\sin ^2}x + \dfrac{1}{2}\sin x = 0\\
\Leftrightarrow \sin x\left( {\sin x + \dfrac{1}{2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = – \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{ – \pi }}{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
2.\cos 3x + \cos 2x – \cos x – 1 = 0\\
\Leftrightarrow \left( {\cos 3x – \cos x} \right) + \left( {\cos 2x – 1} \right) = 0\\
\Leftrightarrow – 2\sin 2x\sin x + \left( {1 – 2{{\sin }^2}x – 1} \right) = 0\\
\Leftrightarrow – 2\sin 2x\sin x – 2{\sin ^2}x = 0\\
\Leftrightarrow – 2\sin x\left( {\sin 2x + \sin x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin 2x = – \sin x = \sin \left( { – x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2x = – x + k2\pi \\
2x = \pi + x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
3x = k2\pi \\
x = \pi + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
\end{array}$