pt $<=>\frac{1+cos6x}{2}.cos2x-\frac{1+cos2x}{2} = 0$$<=> (1 +cos6x)cos2x - (1+cos2x) = 0$$<=>(1+4cos^{3}2x-3cos2x)cos2x-(1+cos2x)=0$$<=>4cos^{4}2x-3cos^{2}2x-1=0$$=> cos2x =1$ hoặc $cos2x=\frac{-1}{4}$(loại)$cos^{2}2x = 1< => \frac{1+cos4x}{2} = 1<=>cos4x=1$tự giải nốt nhá
pt $<=>\frac{1+cos6x}{2}.cos2x-\frac{1+cos2x}{2} = 0$$<=> (1 +cos6x)cos2x - (1+cos2x) = 0$$<=>(1+4cos^{3}2x-3cos2x)cos2x-(1+cos2x)=0$$<=>4cos^{4}2x-3cos^{2}2x-1=0$$=> cos2x =1$ hoặc $cos2x=\frac{-1}{4}$(loại)$cos2x = 1 => 1 - 2sin^{2}x = 1<=>sinx=0$tự giải quyết nốt nhá
pt $<=>\frac{1+cos6x}{2}.cos2x-\frac{1+cos2x}{2} = 0$$<=> (1 +cos6x)cos2x - (1+cos2x) = 0$$<=>(1+4cos^{3}2x-3cos2x)cos2x-(1+cos2x)=0$$<=>4cos^{4}2x-3cos^{2}2x-1=0$$=> cos2x =1$ hoặc $cos2x=\frac{-1}{4}$(loại)$cos
^{2}2x = 1
< =>
\frac{1
+cos
4x}{2} = 1<=>
cos
4x=
1$tự giải nốt nhá