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nastua651
@nastua651
July 2022
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Verified answer
Cos²x+sinxcosx=1 [-π;π]
sinxcosx-(1-cos²x)=0
sinxcosx-sin²x=0
sinx(cosx-sinx)=0
sinx=0 или cosx-sinx=0 |:cosx≠0
x=πn, n∈Z 1-tgx=0
tgx=1
x=π/4+πn, n∈Z
x∈[-π;π]
x₁=-π
x₂=-π+π/4=-3π/4
x₃=0
x₄=π/4
x₅=π
(x₁+x₂+x₃+x₄+x₅):5=(-π -3π/4 +0 +π/4 +π):5=(-π/2):5=-π/10
Ответ: -π/10
Ответ: 3 корня
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Answers & Comments
Verified answer
Cos²x+sinxcosx=1 [-π;π]sinxcosx-(1-cos²x)=0
sinxcosx-sin²x=0
sinx(cosx-sinx)=0
sinx=0 или cosx-sinx=0 |:cosx≠0
x=πn, n∈Z 1-tgx=0
tgx=1
x=π/4+πn, n∈Z
x∈[-π;π]
x₁=-π
x₂=-π+π/4=-3π/4
x₃=0
x₄=π/4
x₅=π
(x₁+x₂+x₃+x₄+x₅):5=(-π -3π/4 +0 +π/4 +π):5=(-π/2):5=-π/10
Ответ: -π/10
Ответ: 3 корня