1 bai luong giac

P

passingby

Chỗ kia là cos3x hả bạn ? :-s
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H

heartrock_159

[TEX]pt \Leftrightarrow (sin2x+cos2x)^3 - 3sin2x.cos2x(sin2x+cos2x+1) + 1 = 0 [/TEX]
Đặt [TEX]t = sin2x + cos2x = \sqrt2cos(2x-\frac{\pi}4) \Rightarrow -\sqrt2 \leq t \leq \sqrt2[/TEX]
[TEX]sin2x.cos2x = \frac{t^2-1}2[/TEX]
[TEX]pt \Leftrightarrow t^2 + 3t^2 - 3t -5 = 0 [/TEX]
[TEX]\Leftrightarrow (t+1)(t^2 + 2t -5) = 0[/TEX]
[TEX]\Leftrightarrow {\[ {t = -1} \\ { t = -1 \pm \sqrt6 \ \ (loai)}[/TEX]
[TEX]\Leftrightarrow t = -1[/TEX]
[TEX]\Leftrightarrow cos{\left(2x + \frac{\pi}4 \right) = cos{\left( \frac{3\pi}4 \right)[/TEX]
 
H

heartrock_159

[TEX]pt \Leftrightarrow (sin2x+cos2x)^3 - 3sin2x.cos2x(sin2x+cos2x+1) + 1 = 0 [/TEX]
Đặt [TEX]t = sin2x + cos2x = \sqrt2cos(2x-\frac{\pi}4) \Rightarrow -\sqrt2 \leq t \leq \sqrt2[/TEX]
[TEX]sin2x.cos2x = \frac{t^2-1}2[/TEX]
[TEX]pt \Leftrightarrow t^2 + 3t^2 - 3t -5 = 0 [/TEX]
[TEX]\Leftrightarrow (t+1)(t^2 + 2t -5) = 0[/TEX]
[TEX]\Leftrightarrow {\[ {t = -1} \\ { t = -1 \pm \sqrt6 \ \ (loai)}[/TEX]
[TEX]\Leftrightarrow t = -1[/TEX]
[TEX]\Leftrightarrow cos{\left(2x + \frac{\pi}4 \right) = cos{\left( \frac{3\pi}4 \right)[/TEX]
 
2

211219942009

hic hic minh cung nghi rui n ma van k ra
bai này trong sách của thầy trần phương ma tớ nháp mãi k ra bạn ak
 
H

heartrock_159

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