1 bài LG hóc búa ạ

N

nguyenbahiep1

[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
 
T

tuyentuyen12a1

mình lại có cách làm khác
mình sẽ bién đỏi như thé này nè
sin3x.cos^3x + sin^3x.cos3x = 3/4.sin^4x
<=> sin3x.cosx.cos^2x + sinx.cos3x.sin^2x = 3/4.sin^4x
<=>sin3x.cosx.(1+cos2x)/2 + sinx.cos3x.(1-cos2x)/2=3/4.sin^4x
<=>sin3x.cosx+sinx.cos3x+cos2x(sin3xcosx-sinxcos3x)=3/2(sin^4x)
<=>sin4x+cos2xsin2x=3/2(sin^4x)
<=>2sin4x+2cos2xsin2x=3sin^4x
<=>2sin4x+sin4x=3sin^4x
<=>sin4x=sin^4x
tới đây bạn giải giống bạn gì đó nha!
 
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