[TEX][I]1 + sin^3x + cos^3x = \frac{3}{2}sin2x[/I][/TEX]
[TEX]\leftrightarrow sin^3x + cos^3x = 3sin2x - 1[/TEX]
[TEX]\leftrightarrow (sinx + cosx).(1 - sinx.cosx) = 3sinx.cosx - 1 (1)[/TEX]
Đặt [TEX]t = sinx + cosx = \sqrt{2}sin(x + \frac{\pi}{4}) \rightarrow t \in [-\sqrt{2} ; \sqrt{2}][/TEX]
[TEX] \rightarrow sinx.cosx = \frac{t^2 - 1}{2}[/TEX]
[TEX]\rightarrow (1) \leftrightarrow t.(1 - \frac{t^2 - 1}{2}) = 3.\frac{t^2 - 1}{2} - 1[/TEX]
[TEX]\leftrightarrow t^3 + 3t^2 - 3t - 5 = 0 [/TEX]
Giải tìm t rồi thế vào [TEX] sinx + cosx[/TEX]
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