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VasyanDoctor
@VasyanDoctor
July 2022
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Решить неравенство:
a) Sin(x) > корень из 2/2
б) 2Cos(x) - 1 > 0
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kirichekov
Verified answer
Sinx>√2/2
x∈(π/4+2πn;3π/4+2πn), n∈Z
2cosx-1>0
cosx>1/2
x∈(-π/3+2πn;π/3+2πn), n∈Z
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Answers & Comments
Verified answer
Sinx>√2/2x∈(π/4+2πn;3π/4+2πn), n∈Z
2cosx-1>0
cosx>1/2
x∈(-π/3+2πn;π/3+2πn), n∈Z