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Cra2y
@Cra2y
July 2022
1
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Решите тригонометрическое уравнение: [tex] \sqrt{3} cos ^{2} x -0.5sin2x=0
[/tex]
[tex]cos ^{2}x+sin2x=0
[/tex]
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sedinalana
Verified answer
1
√3cos²x-sinxcosx=0
cosx(√3cosx-sinx)=0
cosx=0⇒x=π/2+πn,n∈z
√3cosx-sinx=0/cosx
√3-tgx=0
tgx=√3⇒x=π/3+πk,k∈z
2
cos²x+2sinxcosx=0
cosx(cosx+2sinx)=0
cosx+2sinx=0/cosx
1+2tgx=0
tgx=-1/2⇒x=-arctg1/2+πk,k∈z
cosx=0⇒x=π/2+πn,n∈z
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Answers & Comments
Verified answer
1√3cos²x-sinxcosx=0
cosx(√3cosx-sinx)=0
cosx=0⇒x=π/2+πn,n∈z
√3cosx-sinx=0/cosx
√3-tgx=0
tgx=√3⇒x=π/3+πk,k∈z
2
cos²x+2sinxcosx=0
cosx(cosx+2sinx)=0
cosx+2sinx=0/cosx
1+2tgx=0
tgx=-1/2⇒x=-arctg1/2+πk,k∈z
cosx=0⇒x=π/2+πn,n∈z