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Trigonometric equations

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data

    The general solution of sinx.siny=1 is

    1.x=n(pi)+(pi/2)(-1)^n+1 y=m(pi)+(pi/2)(-1)^m+1
    2.x=(4n+1)(pi/2) y=(4m+1)(pi/2)
    3.x=(4n-1)(pi/2) y=(4m-1)(pi/2)

    where n,m belong to I
    2. Relevant equations
    Answer- 2 and 3

    3. The attempt at a solution

    This is possible only if sinx=siny=1 or sinx=siny= -1

    Substituting arbitrary values of n,m in 1st choice.....we get correct answer
    but 1st choice is not included in correct answers
    WHY?
     
  2. jcsd
  3. Sep 1, 2010 #2

    Mentallic

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    Homework Helper

    Frankly, I don't see the point of having y=f(n) since y=x.
    The first doesn't work because you can choose n and m such that sin(x)=-sin(y) so sin(x)sin(y)=-1

    The general solutions are [tex]x= \pi/2 +2\pi n[/tex] such that sin(x)=1, and [tex]y=-\pi /2 +2\pi n[/tex] such that sin(y)=-1. These are equivalent to your second and third equation.

    [tex]z=\pi/2+\pi n[/tex] gives sin(z)=[itex]\pm[/itex]1
     
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