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July 2022
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sin х-cosх>1............................................
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avilov41
Sinx-cosx=√2(1/√2*sinx-1/√2cosx)=√2sin(x-π/4)
Решим неравенство √2sin(x-π/4)>1.
sin(x-π/4)>1/√2
π/4+2πn<x-π/4<3π/4+2πn,n∈z или
π/2+2πn<x<π+2πn,n∈z.
Ответ: π/2+2πn<x<π+2πn,n∈z.
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Answers & Comments
Решим неравенство √2sin(x-π/4)>1.
sin(x-π/4)>1/√2
π/4+2πn<x-π/4<3π/4+2πn,n∈z или
π/2+2πn<x<π+2πn,n∈z.
Ответ: π/2+2πn<x<π+2πn,n∈z.