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Omgg
@Omgg
July 2022
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Решите уравнение
6sin^2x-5cosx-5=0
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русская25
6sin²x-5cosx-5=0
6(1-cos²x)-5cosx-5=0
6-6cos²x-5cosx-5=0
-6cos²x-5cosx+1=0 |*(-1)
6cos²x+5cosx-1=0
Обозначим: cosx=t, тогда
6t²+5t-1=0
D= 25+24=49
t₁= -5+7/12 = 2/12= 1/6
t₂=-5-7/12 = -12/12= -1
1)cosx=1/6
x= +- arccos 1/6 + 2πn,n∈z
2)cosx=-1
x=π+2πn,n∈z
2 votes
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Answers & Comments
6(1-cos²x)-5cosx-5=0
6-6cos²x-5cosx-5=0
-6cos²x-5cosx+1=0 |*(-1)
6cos²x+5cosx-1=0
Обозначим: cosx=t, тогда
6t²+5t-1=0
D= 25+24=49
t₁= -5+7/12 = 2/12= 1/6
t₂=-5-7/12 = -12/12= -1
1)cosx=1/6
x= +- arccos 1/6 + 2πn,n∈z
2)cosx=-1
x=π+2πn,n∈z