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hisdetka
@hisdetka
July 2022
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Решите уравнение: sin^{2} x-3sin x+1=0
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sedinalana
Verified answer
Sinx=a
a²-3a+1=0
D=9-4=5
a1=(3-√5)/2⇒sinx=(3-√5)/2⇒x=(-1)^n*arcsin[(3-√5)/2]+πn,n∈z
a2=(3+√5)/2⇒sinx=(3+√5)/2>1 нет решения
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Answers & Comments
Verified answer
Sinx=aa²-3a+1=0
D=9-4=5
a1=(3-√5)/2⇒sinx=(3-√5)/2⇒x=(-1)^n*arcsin[(3-√5)/2]+πn,n∈z
a2=(3+√5)/2⇒sinx=(3+√5)/2>1 нет решения