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ЧернаяMamba
@ЧернаяMamba
July 2022
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помогите решить уравнение 2cos^2 x + sin2x = sin(x -3п/2) - cos(п/2 +x)
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sedinalana
Verified answer
Sin(3π/2-a)=-cosa ⇒sin(x-3π/2)=-sin(3π/2-x)=cosx
cos(π/2+x)=-sinx
sin2x=2sinx*cosx
2cos²x+2sinxcosx=cosx+sinx
2cosx*(cosx+sinx)-(cosx+sinx)=0
(cosx+sinx)*(2cosx-1)=0
cosx+sinx=0/cosx
1+tgx=0
tgx=-1
x=-π/4+πn,n∈z
2cosx-1=0
2cosx=1
cosx=1/2
x=+-π/3+2πk,k∈z
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Answers & Comments
Verified answer
Sin(3π/2-a)=-cosa ⇒sin(x-3π/2)=-sin(3π/2-x)=cosxcos(π/2+x)=-sinx
sin2x=2sinx*cosx
2cos²x+2sinxcosx=cosx+sinx
2cosx*(cosx+sinx)-(cosx+sinx)=0
(cosx+sinx)*(2cosx-1)=0
cosx+sinx=0/cosx
1+tgx=0
tgx=-1
x=-π/4+πn,n∈z
2cosx-1=0
2cosx=1
cosx=1/2
x=+-π/3+2πk,k∈z