[laTEX]4 tan 2x - 4tan 3x = tan 2x + tan^23x.tan 2x \\ \\dk: cos2x. cos3x \not = 0 \\\\ -4.\frac{sin x}{cos2x.cos3x} = tan 2x. (1 + tan^23x) \\ \\-4.\frac{sin x}{cos2x.cos3x} = \frac{2sin x.cosx}{cos2x.cos^23x} \\ \\TH_1: sin x = 0 \\ \\TH_2 : -2 = \frac{cosx}{cos3x} \\\\ 6cosx - 8cos^3x = cosx [/laTEX]