Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}{\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x + {\cos ^2}4x = 2\\ \Leftrightarrow \dfrac{{1 + \cos 2x}}{2} + \dfrac{{1 + \cos 4x}}{2} + \dfrac{{1 + \cos 6x}}{2} + \dfrac{{1 + \cos 8x}}{2} = 2\\ \Leftrightarrow \cos 2x + \cos 4x + \cos 6x + \cos 8x = 0\\ \Leftrightarrow 2\cos 5x\cos 3x + 2\cos 5x\cos x = 0\\ \Leftrightarrow 2\cos 5x\left( {\cos 3x + \cos x} \right) = 0\\ \Leftrightarrow 4\cos 5x\cos 2x\cos x = 0\\ \Leftrightarrow \cos x\cos 2x\cos 5x = 0\end{array}\)
`⇔` \(\left[ \begin{array}{l}cos x=0\\cos 2x=0\\ cos 5x=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\ (k \in \mathbb{Z})\\ x=\dfrac{\pi}{10}+k\dfrac{\pi}{5}\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy ............
