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Mirumri
@Mirumri
May 2021
1
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Неравенства
2sin*3x>0
и второе 2sin*3x<0
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agasikkk
Sinx>0⇒x∈(2πn;π+2πn)
sinx<0⇒x∈(π+2πn;2π+2πn)
1)
2sin3x>0
sin3x>0
3x∈(2πn;π+2πn)
x∈(2πn/3;π/3+2πn/3)
2)
2sin3x<0
sin3x<0
3x∈(π+2πn;2π+2πn)
x∈(π/3+2πn/3;2π/3+2πn/3)
2 votes
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Answers & Comments
sinx<0⇒x∈(π+2πn;2π+2πn)
1)
2sin3x>0
sin3x>0
3x∈(2πn;π+2πn)
x∈(2πn/3;π/3+2πn/3)
2)
2sin3x<0
sin3x<0
3x∈(π+2πn;2π+2πn)
x∈(π/3+2πn/3;2π/3+2πn/3)