$cos2x+cos(1-4x)=0$
$⇔cos2x=-cos(1-4x)$
$⇔cos2x=cos[\pi-(1-4x)]$
$⇔cos2x=cos(\pi+4x-1)$
$⇔$\(\left[ \begin{array}{l}2x=\pi+4x-1+k2\pi\\2x=-\pi-4x+1+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}-2x=\pi-1+k2\pi\\6x=-\pi+1+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{2}+\dfrac{1}{2}-k\pi\\x=-\dfrac{\pi}{6}+\dfrac{1}{6}+\dfrac{k\pi}{3}\end{array} \right.\)
