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Olli27
@Olli27
July 2022
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3sin²2x+7cos2x-3>=0
Помогите срочно решить, пожалуйста!
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sedinalana
Verified answer
3-3сos²2x+7cos2x-3≥0
cos2x=a
3a²-7a≥0
a(3a-7)≥0
a=0 a=7/3
a≤0⇒cos2x≤0⇒π/2+2πn≤2x≤3π/2+2πn⇒x∈[π/4+πn;3π/4+πn,n∈z]
a≥7/3⇒cos2x≥7/3>1 нет решения
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Olli27
Спасибо большое!!!
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Answers & Comments
Verified answer
3-3сos²2x+7cos2x-3≥0cos2x=a
3a²-7a≥0
a(3a-7)≥0
a=0 a=7/3
a≤0⇒cos2x≤0⇒π/2+2πn≤2x≤3π/2+2πn⇒x∈[π/4+πn;3π/4+πn,n∈z]
a≥7/3⇒cos2x≥7/3>1 нет решения