solve for sin7x =sin3x give general solutions in radians?
回答 (6)
2016-03-22 10:40 am
sin7x =sin3x
7x = 2nπ + 3x or 7x = 2nπ + (π - 3x)
4x = 2nπ or 10x = (2n + 1)π
x = nπ/2 or x = (2n + 1)π/10 ...... where n ∈ Z
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Note : Wayne DeguMan's method is too complicated and the answers can be further simiplied.
x = ±(π/10) + (2nπ/5) is equivalent to x = (2n + 1)π/10
[x = nπ or x = (π/2) + nπ] is equivalent to x = nπ/2
7x = 2nπ + 3x or 7x = 2nπ + (π - 3x)
4x = 2nπ or 10x = (2n + 1)π
x = nπ/2 or x = (2n + 1)π/10 ...... where n ∈ Z
====
Note : Wayne DeguMan's method is too complicated and the answers can be further simiplied.
x = ±(π/10) + (2nπ/5) is equivalent to x = (2n + 1)π/10
[x = nπ or x = (π/2) + nπ] is equivalent to x = nπ/2
2016-03-22 9:33 am
sinA - sinB ≡ 2cos[(A + B)/2]sin[(A - B)/2]
Hence, with A => 7x and B => 3x we have:
sin7x - sin3x ≡ 2cos5xsin2x
Hence, if sin7x - sin3x = 0 then,
2cos5xsin2x = 0
so, either cos5x = 0 or sin2x = 0
Then, 5x = ±π/2 + 2nπ or 2x = 2nπ or 2x = π + 2nπ
i.e. x = ±π/10 + 2nπ/5 or x = nπ or x = π/2 + nπ...for n ∈ ℤ
:)>
Hence, with A => 7x and B => 3x we have:
sin7x - sin3x ≡ 2cos5xsin2x
Hence, if sin7x - sin3x = 0 then,
2cos5xsin2x = 0
so, either cos5x = 0 or sin2x = 0
Then, 5x = ±π/2 + 2nπ or 2x = 2nπ or 2x = π + 2nπ
i.e. x = ±π/10 + 2nπ/5 or x = nπ or x = π/2 + nπ...for n ∈ ℤ
:)>
2016-11-11 10:08 pm
sin7x =sin3x
7x = 2nπ + 3x or 7x = 2nπ + (π - 3x)
4x = 2nπ or 10x = (2n + 1)π
x = nπ/2 or x = (2n + 1)π/10 ................................................................................................................................. where n ∈ z
===%3
7x = 2nπ + 3x or 7x = 2nπ + (π - 3x)
4x = 2nπ or 10x = (2n + 1)π
x = nπ/2 or x = (2n + 1)π/10 ................................................................................................................................. where n ∈ z
===%3
2016-08-02 12:55 pm
sin7x =sin3x
7x = 2nπ + 3x or 7x = 2nπ + (π - 3x)
4x = 2nπ or 10x = (2n + 1)π
x = nπ/2 or x = (2n + 1)π/10 ................................................................................................................................. where n ∈ z
===%3
7x = 2nπ + 3x or 7x = 2nπ + (π - 3x)
4x = 2nπ or 10x = (2n + 1)π
x = nπ/2 or x = (2n + 1)π/10 ................................................................................................................................. where n ∈ z
===%3
2016-03-24 7:41 am
x=3π/2
2016-03-22 10:57 am
sin7x-sin3x=0 and using
sinA-sinB=2sin[(A-B)/2]cos[(A+B)/2]
2sin2xcos5x=0 so
sin2x=0 giving 2x=npi so x=npi/2
or cos5x=0 so 5x=(2r+1)pi/2 so x=(2r+1)pi/2
where n and r are integers, +,- or 0.
sinA-sinB=2sin[(A-B)/2]cos[(A+B)/2]
2sin2xcos5x=0 so
sin2x=0 giving 2x=npi so x=npi/2
or cos5x=0 so 5x=(2r+1)pi/2 so x=(2r+1)pi/2
where n and r are integers, +,- or 0.
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