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12qwaszx0987654321
@12qwaszx0987654321
July 2022
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Решите уравнение пожалуйста 1/sin^2x -3/sinx +2=0
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sedinalana
Verified answer
Sinx=a
1/a²-3/a+2=0
2a²-3a+1=0,a≠0
D=9-8=1
a1=(3-1)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn,n∈z
a2=(3+1)/4=1⇒sinx=1⇒x=π/2+2πk,k∈z
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Answers & Comments
Verified answer
Sinx=a1/a²-3/a+2=0
2a²-3a+1=0,a≠0
D=9-8=1
a1=(3-1)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn,n∈z
a2=(3+1)/4=1⇒sinx=1⇒x=π/2+2πk,k∈z