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namotrasnik
@namotrasnik
July 2022
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Помогите, пожалуйста, решить данный пример
[tex]sinx \geq cos2x[/tex]
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sedinalana
Verified answer
Sinx-1+2sin²x≥0
sinx=a
2a²+a-1≥0
D=1+8=9
a1=(-1-3)/4=-1 U a2=(-1+3)/4=1/2
a≤-1⇒sinx≤-1⇒sinx=-1⇒x=-π/2+2πn
a≥1/2⇒sinx≥1/2⇒x∈[π/6+2πn;5π/6+2πn]
Ответ x∈[π/6+2πn;5π/6+2πn] U {-π/2+2πn},n∈z
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Answers & Comments
Verified answer
Sinx-1+2sin²x≥0sinx=a
2a²+a-1≥0
D=1+8=9
a1=(-1-3)/4=-1 U a2=(-1+3)/4=1/2
a≤-1⇒sinx≤-1⇒sinx=-1⇒x=-π/2+2πn
a≥1/2⇒sinx≥1/2⇒x∈[π/6+2πn;5π/6+2πn]
Ответ x∈[π/6+2πn;5π/6+2πn] U {-π/2+2πn},n∈z