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alisha15sherban
@alisha15sherban
August 2022
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Решить уравнение y' = 0, если y = 2cosx-(5-2x)sinx+4
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sharadi
Y'= - 2Sinx - ( (-2)Sinx +(5 -2x)*(Cosx) ) = -2Sinx -(-2Sinx +5Cosx -2xCosx)=
=-2Sinx +2Sinx -5Cosx +2xCosx = -5Cosx +2xCosx
по условию y' =0
-5Cosx +2xCosx = 0
Сosx(-5 + 2x) = 0
Cosx = 0 или -5 + 2х = 0
x = π/2+πk , k ∈Z
2x = 5
x = 2,5
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Answers & Comments
=-2Sinx +2Sinx -5Cosx +2xCosx = -5Cosx +2xCosx
по условию y' =0
-5Cosx +2xCosx = 0
Сosx(-5 + 2x) = 0
Cosx = 0 или -5 + 2х = 0
x = π/2+πk , k ∈Z 2x = 5
x = 2,5