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Lizo4ка
@Lizo4ка
July 2022
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а) решите уравнение 1+ctg 2x = 1/cos (3п/2-2x)
б) Укажите корни этого уравнения, принадлежащие промежутку [-2ПИ; -ПИ/2]
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1+ctg2x=1/cos(3π/2-2x)
1+ctg2x=1/-sin2x
1+(cos2x/sin2x)=1/-sin2x
1+(cos2x/sin2x)+(1/sin2x)=0
(cos2x+1)/sin2x=-1
cos2x+1=-sin2x
cos²x-sin²x+sin²x+cos²x-2sinxcosx=0
2cos²x-2sinxcosx=0
2cosx(cosx-sinx)=0
cosx=0 или cosx-sinx=0|:cosx≠0
x=π/2+πn, n∈Z 1-tgx=0
tgx=1
x=π/4+πk, k∈Z
На [-2π; -π/2]
x={-7π/4; -3π/2; -3π/4;-π/2}
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Answers & Comments
Verified answer
1+ctg2x=1/cos(3π/2-2x)1+ctg2x=1/-sin2x
1+(cos2x/sin2x)=1/-sin2x
1+(cos2x/sin2x)+(1/sin2x)=0
(cos2x+1)/sin2x=-1
cos2x+1=-sin2x
cos²x-sin²x+sin²x+cos²x-2sinxcosx=0
2cos²x-2sinxcosx=0
2cosx(cosx-sinx)=0
cosx=0 или cosx-sinx=0|:cosx≠0
x=π/2+πn, n∈Z 1-tgx=0
tgx=1
x=π/4+πk, k∈Z
На [-2π; -π/2]
x={-7π/4; -3π/2; -3π/4;-π/2}