sin^3x - cos^3x = sinx + cosx
2sin^3x - cos2x + cosx = 0
[TEX]a^3 - b^3 = (a-b)(a^2 + ab + b^2)[/TEX]
[TEX]pt \Leftrightarrow (sinx-cosx)(1+sinx.cosx) = sinx + cosx[/TEX]
[TEX]\Leftrightarrow 2cosx = (sinx-cosx).sinx.cosx [/TEX]
[TEX]\Leftrightarrow cosx(2-sin^2x - sinx.cosx)=0[/TEX]
[TEX]\Leftrightarrow {\[ {cosx = 0} \\ {sin^2x - sinx.cosx + 2cos^2x = 0 \ \ (2)}[/TEX]
Với [TEX]cosx \not= 0 [/TEX] thì (2) tương đương
[TEX]tan^2x - tanx + 2 = (tanx - \frac12) + \frac74 = 0 \ \ (VN)[/TEX]
Vậy [TEX]x = \frac{\pi}2 + k \pi , \ k \in \mathbb{Z}[/TEX]