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WhiteKing21
@WhiteKing21
July 2022
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2sin2x>-√(2), Якщо можна то з розвязком нерівності на одиничному колі і з поясненями)))
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sedinalana
Verified answer
2sin2x>-√2
sin2x=-√2/2
-π/4+2πn≤2x≤5π/4+2πn,n∈z
-π/8+πn≤x≤8π/8+πn,n∈z
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skyne8
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Sin2x>-√2/2, 5π/4 +2πk>2x>-π/4+2πk, 5π/8+πk>x>-π/8+πk, k∈Z
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Answers & Comments
Verified answer
2sin2x>-√2sin2x=-√2/2
-π/4+2πn≤2x≤5π/4+2πn,n∈z
-π/8+πn≤x≤8π/8+πn,n∈z
Verified answer
Sin2x>-√2/2, 5π/4 +2πk>2x>-π/4+2πk, 5π/8+πk>x>-π/8+πk, k∈Z