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Katyusha5
@Katyusha5
July 2022
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2cos квадрат x+5cosx=3
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русская25
2cos²x+5cosx=3
2cos²x+5cosx-3=0
Замена: cosx=t
2t²+5t-3=0
D= 25+4*2*3 = 25+24 = 49
t₁= (-5+7)/4 = 2/4 = 1/2
t₂= (-5-7)/4 = -12/4 = -3
cosx= 1/2
x= +-arccos1/2+2πk
x₁= +- π/3+2πk,k∈z
cosx=-3 -нет корней
Ответ: +- π/3+2πk,k∈z
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Answers & Comments
2cos²x+5cosx-3=0
Замена: cosx=t
2t²+5t-3=0
D= 25+4*2*3 = 25+24 = 49
t₁= (-5+7)/4 = 2/4 = 1/2
t₂= (-5-7)/4 = -12/4 = -3
cosx= 1/2
x= +-arccos1/2+2πk
x₁= +- π/3+2πk,k∈z
cosx=-3 -нет корней
Ответ: +- π/3+2πk,k∈z