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Math Help - find all solutions

  1. #1
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    find all solutions

    hey guys I am a little confused on where to start! please help?
    so find all solutions to this equation by using graphs...
    sinx=sin(1/2x)
    Last edited by caseygaspar; April 30th 2009 at 09:43 AM.
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  2. #2
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    Quote Originally Posted by caseygaspar View Post
    hey guys I am a little confused on where to start! please help?
    so find all solutions to this equation by using graphs...
    sinx=sin(1/2x)
     \sin(K) = \sin(P) \rightarrow K = \arcsin(\sin(P))

    And  \arcsin(\sin(P)) = P
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by caseygaspar View Post
    hey guys I am a little confused on where to start! please help?
    so find all solutions to this equation by using graphs...
    sinx=sin(1/2x)
    To find solutions using graphs, you simply graph each side of the equation as a function: y = sin(x) and y=sin(1/2x). The solutions are the x-values at each point where the two graphs intersect.
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    In these types of problems, you'll want all of your trig functions to be evaluated at the same number (in this case you've got x AND \frac{x}{2}).

    Use the half-angle identity for sine: \sin(\frac{x}{2})={\pm}\sqrt{\frac{1-\cos(x)}{2}}

    So this root equals \sin(x)

    Square both sides and you'll have \frac{1-\cos(x)}{2}=\sin(x)^2

    Then use the pythagorean identity: \sin(x)^2=1-\cos(x)^2

    You now have \frac{1-\cos(x)}{2}=1-\cos(x)^2

    Put this equation into standard form for a quadratic equation and let u=\cos(x)

    You now have 0=2u^2-u-1

    Factor, solve, and put cosine back in for u to get: \cos(x)=-\frac{1}{2} and \cos(x)=1

    Thus, all solutions are x=\arccos(-\frac{1}{2}) and x=\arccos(1)

    Don't forget that the range of the arccosine function is limited, so there are three solutions to the original question in every period of the cosine.
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  5. #5
    Member pflo's Avatar
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    Quote Originally Posted by sinewave85 View Post
    To find solutions using graphs, you simply graph each side of the equation as a function: y = sin(x) and y=sin(1/2x). The solutions are the x-values at each point where the two graphs intersect.
    You could also get the equation equal to zero: \sin(x)-\sin(\frac{x}{2})=0 and graph this as a function. Then the zeros of the function are the solutions (the x-intercepts of y=\sin(x)-\sin(\frac{x}{2})).
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  6. #6
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    Quote Originally Posted by Mush View Post
     \sin(K) = \sin(P) \rightarrow K = \arcsin(\sin(P))

    And  \arcsin(\sin(P)) = P
    Actually it's  \sin(K) = \sin(P) \Rightarrow K = \arcsin(\sin(P)) + 2 n \pi or K = [\pi - \arcsin(\sin(P))] + 2 n \pi.
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