Phương trình đã cho tương đương với:
\(cos2x+\left(cos6x+cos10x\right)=0\)
\(\Leftrightarrow cos2x+2.cos8x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2cos8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\1+2cos8x=0\end{matrix}\right.\)
+ TH1:
\(cos2x=0\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\left(k\in Z\right)\)
+ TH2:
\(1+2cos8x=0\Leftrightarrow cos8x=-\dfrac{1}{2}=cos\dfrac{2\pi}{3}\)
\(\Leftrightarrow8x=\pm\dfrac{2\pi}{3}+k2\pi\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{4}\end{matrix}\right.\) \(\left(k\in Z\right)\)
Vậy phương trình gồm các họ nghiệm: \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\), \(x=\dfrac{\pi}{12}+\dfrac{k\pi}{4}\), \(x=-\dfrac{\pi}{12}+\dfrac{k\pi}{4}\) với \(k\in Z\)