cosx + cos3x + 2cos5x=0<=>$4cos^{3}x -2cosx + 2cos(3x+2x)=0$<=>$2cosx(2cos^{2}-1)+2cos3xcos2x- 2sin3xsin2x=0$<=>$cosxcos2x+cos3xcos2x-sin3xsin2x=0$<=>$cos2x(cosx +cos3x)-sin3xsin2x=0$<=>$2cos^{2}2xcosx- 2sinxcosxsin3x=0$<=> $2cosx(cos^{2}2x- sinxsin3x)= 0$<=>$\begin{cases}cosx=0 <=> x=\frac{\pi }{2}+k \pi \\ cos^{2}2x -sinxsin3x=0(*)\end{cases} $$(*)<=> cos^{2}2x-\frac{1}{2}(cos2x-cos4x)=0$<=>$cos^{2}2x -\frac{1}{2}cos2x+cos^{2}2x-\frac{1}{2}=0$<=>$4cos^{2}2x - coss2x -1=0$
cosx + cos3x + 2cos5x=0<=>$4cos^{3}x -2cosx + 2cos(3x+2x)=0$<=>$2cosx(2cos^{2}-1)+2cos3xcos2x- 2sin3xsin2x=0$<=>$cosxcos2x+cos3xcos2x-sin3xsin2x=0$<=>$cos2x(cosx +cos3x)-sin3xsin2x=0$<=>$2cos^{2}2xcosx- 2sinxcosxsin3x=0$<=> $2cosx(cos^{2}2x- sinxsin3x)= 0$<=>$\begin{cases}cosx=0 <=> x=\frac{\pi }{2}+k \pi \\ cos^{2}2x -sinxsin3x=0(*)\end{cases} $$(*)<=> cos^{2}2x-\frac{1}{2}(cos2x-cos4x)=0$<=>$cos^{2}2x -\frac{1}{2}cos2x+cos^{2}2x-\frac{1}{2}=0$<=>$4cos^{2}2x - cos2x -1=0$