Đáp án:
${x = k\pi ;x = \dfrac{1}{2}\arccos \left( {\dfrac{{ - 3}}{4}} \right) + k\pi ;x = - \dfrac{1}{2}\arccos \left( {\dfrac{{ - 3}}{4}} \right) + k\pi \left( {k \in Z} \right)}$
Giải thích các bước giải:
$\begin{array}{l}
6)\cos 4x = {\cos ^2}x\\
\Leftrightarrow \cos 4x = \dfrac{{1 + \cos 2x}}{2}\\
\Leftrightarrow 2\cos 4x - \cos 2x - 1 = 0\\
\Leftrightarrow 2\left( {2{{\cos }^2}2x - 1} \right) - \cos 2x - 1 = 0\\
\Leftrightarrow 4{\cos ^2}2x - \cos 2x - 3 = 0\\
\Leftrightarrow \left( {\cos 2x - 1} \right)\left( {4\cos x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 1\\
\cos 2x = \dfrac{{ - 3}}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k2\pi \\
2x = \arccos \left( {\dfrac{{ - 3}}{4}} \right) + k2\pi \\
2x = - \arccos \left( {\dfrac{{ - 3}}{4}} \right) + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{1}{2}\arccos \left( {\dfrac{{ - 3}}{4}} \right) + k\pi \\
x = - \dfrac{1}{2}\arccos \left( {\dfrac{{ - 3}}{4}} \right) + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
Vậy phương trình có các họ nghiệm là:
${x = k\pi ;x = \dfrac{1}{2}\arccos \left( {\dfrac{{ - 3}}{4}} \right) + k\pi ;x = - \dfrac{1}{2}\arccos \left( {\dfrac{{ - 3}}{4}} \right) + k\pi \left( {k \in Z} \right)}$
