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Hilbrain
@Hilbrain
August 2022
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Помогите решить 20 задание
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sedinalana
Verified answer
√(2x-x²)(sin2x+cosx)=0
ОДЗ
2x-x²≥0
x(2-x)≥0
x=0 x=2
x∈[0;2]
sin2x+cosx=0
2sinxcosx+cosx=0
cosx(2sinx+1)=0
cosx=0⇒x=π/2+πn,n∈z +ОДЗ⇒x=π/2
sinx=-1/2⇒x=(-1)^(n+1)π/6+πn,n∈x +ОДЗ⇒нет решения
Ответ x={0;π/2;2}
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Answers & Comments
Verified answer
√(2x-x²)(sin2x+cosx)=0ОДЗ
2x-x²≥0
x(2-x)≥0
x=0 x=2
x∈[0;2]
sin2x+cosx=0
2sinxcosx+cosx=0
cosx(2sinx+1)=0
cosx=0⇒x=π/2+πn,n∈z +ОДЗ⇒x=π/2
sinx=-1/2⇒x=(-1)^(n+1)π/6+πn,n∈x +ОДЗ⇒нет решения
Ответ x={0;π/2;2}