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Gerbel
@Gerbel
July 2022
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АЛГЕБРА!!!
Помогите, пожалуйста, с тригонометрическими уравнениями, прошу!!!
1. [tex]2sin^{2} x-( \sqrt{3} +2)sinx+ \sqrt{3} =0
[/tex]
2. [tex]2-6sinxcosx+cos(5 \pi -4x)=0[/tex]
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sedinalana
Verified answer
Sinx=a
2a²-(√3+2)a+√3=0
D=3+4√3+4-8√3=3-4√3+4=(√3-2)²
√D=2-√3
a1=(√3+2-2+√3)/4=√3/2⇒sinx=√3/2⇒x=(-1)^n*π/3+πn,n∈z
a2=(√3+2+2-√3)/4=1⇒sinx=1⇒x=π/2+2πk,k∈z
2-3sin2x-cos4x=0
2-3sin2x-1+2sin²2x=0
sin2x=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒sin2x=1/2⇒2x=(-1)^n*π/6+πn⇒x=(-1)^n*π/12+πn/2,n∈z
a2=(3+1)/4=1⇒sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z
2 votes
Thanks 0
Gerbel
Спасибо большое за решение! Я хоть поняла, как надо решать :)
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Answers & Comments
Verified answer
Sinx=a2a²-(√3+2)a+√3=0
D=3+4√3+4-8√3=3-4√3+4=(√3-2)²
√D=2-√3
a1=(√3+2-2+√3)/4=√3/2⇒sinx=√3/2⇒x=(-1)^n*π/3+πn,n∈z
a2=(√3+2+2-√3)/4=1⇒sinx=1⇒x=π/2+2πk,k∈z
2-3sin2x-cos4x=0
2-3sin2x-1+2sin²2x=0
sin2x=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒sin2x=1/2⇒2x=(-1)^n*π/6+πn⇒x=(-1)^n*π/12+πn/2,n∈z
a2=(3+1)/4=1⇒sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z