Đáp án: \[\left[ \begin{array}{l}x = k\pi \\x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in Z} \right)\] Giải thích các bước giải: Ta có: \(\begin{array}{l}\cos 3x = 4{\cos ^3}x – 3\cos x\\\cos 2x = 2{\cos ^2}x – 1\\\cos 3x + \cos 2x – \cos x – 1 = 0\\ \Leftrightarrow \left( {4{{\cos }^3}x – 3\cos x} \right) + \left( {2{{\cos }^2}x – 1} \right) – \cos x – 1 = 0\\ \Leftrightarrow 4{\cos ^3}x + 2{\cos ^2}x – 4\cos x – 2 = 0\\ \Leftrightarrow \left( {4{{\cos }^3}x – 4{{\cos }^2}x} \right) + \left( {6{{\cos }^2}x – 6\cos x} \right) + \left( {2\cos x – 2} \right) = 0\\ \Leftrightarrow 4{\cos ^2}x.\left( {\cos x – 1} \right) + 6\cos x\left( {\cos x – 1} \right) + 2\left( {\cos x – 1} \right) = 0\\ \Leftrightarrow \left( {\cos x – 1} \right).\left( {4{{\cos }^2}x + 6\cos x + 2} \right) = 0\\ \Leftrightarrow \left( {\cos x – 1} \right)\left( {2{{\cos }^2}x + 3\cos x + 1} \right) = 0\\ \Leftrightarrow \left( {\cos x – 1} \right)\left( {\cos x + 1} \right)\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 1\\\cos x = – 1\\\cos x = – \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \pi + k2\pi \\x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in Z} \right)\end{array}\) Reply
Đáp án:
\[\left[ \begin{array}{l}
x = k\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 3x = 4{\cos ^3}x – 3\cos x\\
\cos 2x = 2{\cos ^2}x – 1\\
\cos 3x + \cos 2x – \cos x – 1 = 0\\
\Leftrightarrow \left( {4{{\cos }^3}x – 3\cos x} \right) + \left( {2{{\cos }^2}x – 1} \right) – \cos x – 1 = 0\\
\Leftrightarrow 4{\cos ^3}x + 2{\cos ^2}x – 4\cos x – 2 = 0\\
\Leftrightarrow \left( {4{{\cos }^3}x – 4{{\cos }^2}x} \right) + \left( {6{{\cos }^2}x – 6\cos x} \right) + \left( {2\cos x – 2} \right) = 0\\
\Leftrightarrow 4{\cos ^2}x.\left( {\cos x – 1} \right) + 6\cos x\left( {\cos x – 1} \right) + 2\left( {\cos x – 1} \right) = 0\\
\Leftrightarrow \left( {\cos x – 1} \right).\left( {4{{\cos }^2}x + 6\cos x + 2} \right) = 0\\
\Leftrightarrow \left( {\cos x – 1} \right)\left( {2{{\cos }^2}x + 3\cos x + 1} \right) = 0\\
\Leftrightarrow \left( {\cos x – 1} \right)\left( {\cos x + 1} \right)\left( {2\cos x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = – 1\\
\cos x = – \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \pi + k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)