[TEX]sin3x.cos^3x+sin^3x.cos3x = \frac{3}{4}.sin^4x \\ sin3x.(\frac{1}{4}.cos3x+\frac{3}{4}.cosx) + cos3x.(\frac{3}{4}.sin x-\frac{1}{4}.sin3x) = \frac{3}{4}.sin^4x \\ sin3x.cos3x + 3sin3x.cosx + 3cos3x.sinx - cos3x.sin3x = 3sin^4x \\ sin 4x = sin^4x \\ TH_1 : sin x = 0 \\ TH_2 : 4cosx.cos2x = sin^3 x \\ 4cosx(2cos^2x-1) = sin^3x \\ 8cos^3x - 4cosx - sin^3x = 0 [/TEX]
chia cả 2 vế cho [TEX]cos^3x[/TEX] là xong