Đáp án:
\[\left[ \begin{array}{l}
x = \pm \dfrac{1}{2}\arccos \dfrac{6}{7} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x\\
\cos 2x = 2{\cos ^2}x - 1\\
3\cos 4x - {\sin ^2}2x + \cos 2x - 2 = 0\\
\Leftrightarrow 3.\left( {2{{\cos }^2}2x - 1} \right) - \left( {1 - {{\cos }^2}2x} \right) + \cos 2x - 2 = 0\\
\Leftrightarrow 6{\cos ^2}2x - 3 - 1 + {\cos ^2}2x + \cos 2x - 2 = 0\\
\Leftrightarrow 7{\cos ^2}2x + \cos 2x - 6 = 0\\
\Leftrightarrow \left( {7\cos 2x - 6} \right)\left( {\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \dfrac{6}{7}\\
\cos 2x = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \pm \arccos \dfrac{6}{7} + 2k\pi \\
2x = \pi + 2k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{1}{2}\arccos \dfrac{6}{7} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Vậy nghiệm của phương trình đã cho là: \(\left[ \begin{array}{l}
x = \pm \dfrac{1}{2}\arccos \dfrac{6}{7} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
