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stetkhem67
@stetkhem67
July 2022
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Народ, привет всем, помогите пожалуйста с простейшими тригонометрическими уравнениями, даю 39 баллов!
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sedinalana
Verified answer
1
cos2x*cosx-sin2x*sinx=√2/2
cosa*cosb-sina*sinb=cos(a+b)
cos(2x+x)=√2/2
cos3x=√2/2
3x=+-π/4+2πn
x=+-π/12+2πn/3,n∈z
2
sin2x_cosx=0
sin2a=2sinacosa
2sinx*cosa+cosx=0
cosx*(2sinx+1)=0
cosx=0⇒x=π/2+πn,n∈z
2sinx+1=0
2sinx=-1
sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈z
3
1-sin3x=(sinx/2-cosx/2)²
(a-b)²=a²-2ab+b²
1-sin3x=sin²x/2-2sinx/2*cosx/2+cos²x/2
sin²a+cos²a=1
2sina*cosa=sin2a
1-sin3x=1-sinx
sin3x=sinx
sin3x-sinx=0
sina-sinb=2sin[(a-b)/2]*cos[(a+b)/2]
2sinx8cos2x=0
sinx=0⇒x=πn,n∈z
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈z
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Answers & Comments
Verified answer
1cos2x*cosx-sin2x*sinx=√2/2
cosa*cosb-sina*sinb=cos(a+b)
cos(2x+x)=√2/2
cos3x=√2/2
3x=+-π/4+2πn
x=+-π/12+2πn/3,n∈z
2
sin2x_cosx=0
sin2a=2sinacosa
2sinx*cosa+cosx=0
cosx*(2sinx+1)=0
cosx=0⇒x=π/2+πn,n∈z
2sinx+1=0
2sinx=-1
sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈z
3
1-sin3x=(sinx/2-cosx/2)²
(a-b)²=a²-2ab+b²
1-sin3x=sin²x/2-2sinx/2*cosx/2+cos²x/2
sin²a+cos²a=1
2sina*cosa=sin2a
1-sin3x=1-sinx
sin3x=sinx
sin3x-sinx=0
sina-sinb=2sin[(a-b)/2]*cos[(a+b)/2]
2sinx8cos2x=0
sinx=0⇒x=πn,n∈z
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈z