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maxxx2000
@maxxx2000
August 2022
1
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Помогите пожалуйста решить неравенство
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sedinalana
2cos²x+3cosx-2<0
cosx=a
2a²+3a-2<0
D=9+16=25
a1=(-3-5)/4=-2 U a2=(-3+5)/4=1/2
+ _ +
-------------------(-2)-----------------(1/2)----------------
-2<a<1/2
-2<cosx<1/2⇒cosx<1/2
x∈(π/6+2πk;11π/6+2πk,k∈z)
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Answers & Comments
cosx=a
2a²+3a-2<0
D=9+16=25
a1=(-3-5)/4=-2 U a2=(-3+5)/4=1/2
+ _ +
-------------------(-2)-----------------(1/2)----------------
-2<a<1/2
-2<cosx<1/2⇒cosx<1/2
x∈(π/6+2πk;11π/6+2πk,k∈z)