Đáp án:
\[x = \pm \dfrac{\pi }{6} + k\pi \,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 2{\cos ^2}x - 1 \Rightarrow \cos 4x = 2{\cos ^2}2x - 1\\
\cos 4x + 7\cos 2x - 3 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}2x - 1} \right) + 7\cos 2x - 3 = 0\\
\Leftrightarrow 2{\cos ^2}2x + 7\cos 2x - 4 = 0\\
\Leftrightarrow \left( {2\cos 2x - 1} \right)\left( {\cos 2x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \dfrac{1}{2}\\
\cos 2x = - 4
\end{array} \right.\\
- 1 \le \cos 2x \le 1 \Rightarrow \cos 2x = \dfrac{1}{2}\\
\Leftrightarrow 2x = \pm \dfrac{\pi }{3} + k2\pi \\
\Leftrightarrow x = \pm \dfrac{\pi }{6} + k\pi \,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
