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nemocapitan
@nemocapitan
August 2022
2
11
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Решите:
[tex] \frac{cos2x+3 \sqrt{2} sinx-3 }{ \sqrt{cosx} } } =0[/tex]
на промежутке [2п; 6п]
Ответ: а) п/4 + 2Пk
б) 9п/4; 17П/4
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sedinalana
Verified answer
ОДЗ
cosx>0⇒x∈(-π/2+2πn;π/2+2πn,n∈z)
cos2x+3√2sinx-3=0
1-2sin²x+3√2sinx-3=0
sinx=a
2a²-3√2a+2=0
D=18-16=2
a1=(3√2+√2)/4=√2⇒sinx=√2>1 нет решения
a2=(3√2-√2)/4=√2/2⇒sinx=(-1)^n*π/4+πn,n∈z
С учетом ОДЗ
x=π/4+2πn,n∈z
0 votes
Thanks 0
Zuipol
Cosx>0⇒x∈(-π/2+2πn;π/2+2πn,n∈z)
a2=(3√2-√2)/4=√2/2⇒sinx=(-1)^n*π/4+πn,n∈z
x=π/4+2πn,n∈z
==========================
0 votes
Thanks 0
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Answers & Comments
Verified answer
ОДЗcosx>0⇒x∈(-π/2+2πn;π/2+2πn,n∈z)
cos2x+3√2sinx-3=0
1-2sin²x+3√2sinx-3=0
sinx=a
2a²-3√2a+2=0
D=18-16=2
a1=(3√2+√2)/4=√2⇒sinx=√2>1 нет решения
a2=(3√2-√2)/4=√2/2⇒sinx=(-1)^n*π/4+πn,n∈z
С учетом ОДЗ
x=π/4+2πn,n∈z
a2=(3√2-√2)/4=√2/2⇒sinx=(-1)^n*π/4+πn,n∈z
x=π/4+2πn,n∈z
==========================