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Marinamalikova4
@Marinamalikova4
September 2021
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Помогите! !Пожалуйста. Очень надо.
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kmike21
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tgα(1+cos2α)=tgα(1+cos²α-sin²α)=tgα(cos²α+cos²α)=2tgαcos²α= 2sinαcos²α/cosα=2sinαcosα=sin2α
3cos2β+sin²β-cos²β=3cos2β-(cos²β-sin²β)=3cos2β-cos2β=2cos2β
cos6xcos5x+sin6xsin5x=1
cos(6x-5x)=1
cosx=1
Ответ: x=2πn, где n - целое
sin2x-2cosx=0
2sinxcosx-2cosx=0
cosx(sinx-1)=0
cosx₁=0, x₁=π/2 +πn
sinx₂=1
x₂=π/2 +2πn - входит в x₁
Ответ:x=π/2 +πn, где n - целое
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Answers & Comments
Verified answer
tgα(1+cos2α)=tgα(1+cos²α-sin²α)=tgα(cos²α+cos²α)=2tgαcos²α= 2sinαcos²α/cosα=2sinαcosα=sin2α
3cos2β+sin²β-cos²β=3cos2β-(cos²β-sin²β)=3cos2β-cos2β=2cos2β
cos6xcos5x+sin6xsin5x=1
cos(6x-5x)=1
cosx=1
Ответ: x=2πn, где n - целое
sin2x-2cosx=0
2sinxcosx-2cosx=0
cosx(sinx-1)=0
cosx₁=0, x₁=π/2 +πn
sinx₂=1
x₂=π/2 +2πn - входит в x₁
Ответ:x=π/2 +πn, где n - целое