(4cos^3 (x) - 3cos (x) )^2. cos2x - cos^2 (x) = 0cos^2 (x) (4cos^2 (x) -3).cos2x - cos^2 (x) = 0 TH1: cos ^2 (x) = 0 tg đg cos x =0TH2 : (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= 0giải pt bậc 3 ẩn cos2x
$(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