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merci0077
@merci0077
August 2022
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сколько корней имеет уравнение 4sin(x/2)-cos x+1=0 на [0;2П]
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kirichekov
Verified answer
4sin(x/2)-cosx+1=0
cosx=cos(2*(x/2))=1-2*sin²(x/2)
4sin(x/2)-(1-2*sin²(x/2))+1=0
2*sin²(x/2)+4*sin(x/2)=0
2*sin(x/2)*(sin(x/2)+2)=0
2sin(x/2)=0 или sin(x/2)+2=0
sin(x/2)+2=0, sin(x/2)=-2. уравнение не имеет решений, т.к. -2∉[-1;1]
∈Z
x=2πn, n∈Z
n=0. x=2*π*0, x=0
n=1. x=2*π*1, x=2π
ответ:
2 корня
2 votes
Thanks 5
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Answers & Comments
Verified answer
4sin(x/2)-cosx+1=0cosx=cos(2*(x/2))=1-2*sin²(x/2)
4sin(x/2)-(1-2*sin²(x/2))+1=0
2*sin²(x/2)+4*sin(x/2)=0
2*sin(x/2)*(sin(x/2)+2)=0
2sin(x/2)=0 или sin(x/2)+2=0
sin(x/2)+2=0, sin(x/2)=-2. уравнение не имеет решений, т.к. -2∉[-1;1]
x=2πn, n∈Z
n=0. x=2*π*0, x=0
n=1. x=2*π*1, x=2π
ответ: 2 корня