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Bagura1703
@Bagura1703
July 2022
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Дам 40 баллов! а) 2cos^2x+5sinx+1=0 б) укажите корни, принадлежащие отрезку П<а<2П. а-это альфа
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sedinalana
Verified answer
2-2sin²x+5sinx+1=0
sinx=a
2a²-5a-3=0
D=25+24=49
a1=(5-7)/4=-1/2⇒sinx=-1/2⇒x=-π/6+2πn,n∈z Г 7π/6+2πлбл∈я
a2=(5+7)/4=3⇒sinx=3>1 нет решения
π<-π/6+2πn<2π
6<-1+12n<12
7<12n<13
7/12<n<13/12
n=1⇒x=-π/6+2π=11π/6
π<7π/6+2πk<2π
6<7+12k<12
-1<12k<5
-1/12<n<5/12
n=0⇒x=7π/6
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Answers & Comments
Verified answer
2-2sin²x+5sinx+1=0sinx=a
2a²-5a-3=0
D=25+24=49
a1=(5-7)/4=-1/2⇒sinx=-1/2⇒x=-π/6+2πn,n∈z Г 7π/6+2πлбл∈я
a2=(5+7)/4=3⇒sinx=3>1 нет решения
π<-π/6+2πn<2π
6<-1+12n<12
7<12n<13
7/12<n<13/12
n=1⇒x=-π/6+2π=11π/6
π<7π/6+2πk<2π
6<7+12k<12
-1<12k<5
-1/12<n<5/12
n=0⇒x=7π/6