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uchenic4
@uchenic4
August 2022
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Решите неравенство: cos^(2)x-0,5sinx>1
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sedinalana
Verified answer
Cos²x-0,5sinx-1>0
1-sin²x-0,5sinx-1>0
sin²x+0,5sinx<0
sinx=a
a²+0,5a<0
a(a+0,5)<0
a=0 a=-0,5
-0,5<a<0
-0,5<sinx<0
-7π/6+2πn<x<-π+2πn,n∈z U -11π+2πn<x<2πn,n∈z
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Verified answer
Cos²x-0,5sinx-1>01-sin²x-0,5sinx-1>0
sin²x+0,5sinx<0
sinx=a
a²+0,5a<0
a(a+0,5)<0
a=0 a=-0,5
-0,5<a<0
-0,5<sinx<0
-7π/6+2πn<x<-π+2πn,n∈z U -11π+2πn<x<2πn,n∈z