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Solution of an equation

Posted on 2013-11-24
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Last Modified: 2013-12-13
Can someone solve this equation? A colleague gave this. This is not homework

Equation
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Question by:Saqib Husain, Syed
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9 Comments
 
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Expert Comment

by:ozo
ID: 39673365
φ=n*2pi, where n ∈ Z
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Author Comment

by:Saqib Husain, Syed
ID: 39679855
Is there a way to reduce the terms and get an expression for phi?
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Expert Comment

by:ozo
ID: 39679879
Yes.  see above.
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Expert Comment

by:ozo
ID: 39679893
phi = n*2pi, where n is an integer.
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Author Comment

by:Saqib Husain, Syed
ID: 39679917
Ok, but can I have some algebraic steps which can lead me to that result?


φ=n*2pi, where n ∈ Z
I do not understand the & and ; symbols neither do I understand the Z
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Expert Comment

by:ozo
ID: 39679928
http:#a39679893 repeats the solution without the symbols.

For the algebra, let x = sin(phi), rewrite the trig functions in terms of sin, and solve for x, than take arcsin(x)
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Expert Comment

by:BigRat
ID: 39680756
ozo's solution is a trivial solution because the sine of any multiple of 2*pi  is zero, the cosine is 1, and the tangent of half the angle (180) also is zero. So the entire RHS reduces to 1. Thus any integer multiple of 2pi is a solution.

If you multiply out the denominator you'll get the left hand term as a power of the constant e equal to a right hand term which does not involve e in the same degree, that is the tangent expression of phi over 2 does not contain an e term. This means that phi must be given in terms of e and therefore in non-trigometrical terms as well.

I suspect thate there is no simple algebraic solution.
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Accepted Solution

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ozo earned 500 total points
ID: 39682622
I dropped a parenthesis in my derivation, and missed a cos(phi) cancellation, which caused me to omit the odd pi solutions;
phi = n*p where n is an integer

You can reduce all the trig to
(1+s)*(1-es)^e - (1+es)^(e+1) = 0  [for s!=1]
where s=sin(phi)
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Author Closing Comment

by:Saqib Husain, Syed
ID: 39716014
Sorry did not get the chance to try this but it is time to close this now. Thanks for the effort.
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