$(4cos^3 x - 3cos x )^2. cos2x - cos^2 x = 0$
$cos^2 x (4cos^2 x -3).cos2x - cos^2 x = 0 $
TH1: $cos ^2 x = 0 tg đg cos x =0$
TH2 : $(4cos ^2 (x) - 3)^2 .cos2x - 1=0$
$(16cos^4 x - 12cos^2 x + 9) .cos2x - 1 = 0$
$(4(1 + cos 2x)^2 - 6.(cos2x + 1) + 9) .cos2x - 1= 0$
giải pt bậc 3 ẩn cos2x