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plotnikov2009
@plotnikov2009
August 2022
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Найти число решений уравнения
3sin^2x-8cosx=0
принадлежащих отрезку [0; 7pi/2]
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sedinalana
Verified answer
3-3cos²x-8cosx=0
cosx=a
3a²+8a-3=0
D=64+36=100
a1=(-8-10)/6=-3⇒cosx=-3<-1 нет решения
a2=(-8+10)/6=1/3⇒cosx=1/3
x=+-arccos1/3+2πn,n∈z
x=arccos1/3
x=2π-arccos1/3
x=2π+arccos1/3
1 votes
Thanks 3
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Answers & Comments
Verified answer
3-3cos²x-8cosx=0cosx=a
3a²+8a-3=0
D=64+36=100
a1=(-8-10)/6=-3⇒cosx=-3<-1 нет решения
a2=(-8+10)/6=1/3⇒cosx=1/3
x=+-arccos1/3+2πn,n∈z
x=arccos1/3
x=2π-arccos1/3
x=2π+arccos1/3