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Homework Help: Equation solving

  1. Dec 19, 2004 #1
    I came upon an equation similar to 0 = (cos x)^2 - x^2 or even 0 = cos x - x and i don't have a clue how to solve it analytically. I tried taking inverse cosine function on both sides but that still doesn't isolate the x. How would you do it?
     
  2. jcsd
  3. Dec 19, 2004 #2

    dextercioby

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    The first eq.includes the second one as it can easily be factorized:
    [tex] \cos^{2}x-x^{2}=(\cos x+x)(\cos x-x)=0 [/tex]

    The two equations obtained as called by the mathematicians "transcendental equations".They are exactly analytically solvable (sorry fo pushing English grammar :tongue2: ) very rarely,and the 2 written above NEVER.
    It this case an approximative method is useful and the first one that comes to my mind is the graphic one.Simply plot [itex] \cos x [/itex],[itex] x [/itex] & [itex]-x [/itex] and the solutions to your problem will be the intersection points.

    Daniel.
     
  4. Dec 19, 2004 #3
    I've approached my original problem differently and now have an equation of the type Acos x = Bcos((pi)x/C)). I have a problem expanding it however. Is there an identity that expands cos(ax) into cos x + f(x) or (cos x)f(x) ?
     
  5. Dec 19, 2004 #4
    you could use a taylor series to get a good approximation of cos x and then you should be able to solve for x from that.

    that is probably more work than you need to do however, but it is a good place to go if you are desperate.
     
  6. Dec 20, 2004 #5
    so there is no ther way? Is it not possible to find an EXACT solution?
     
  7. Dec 20, 2004 #6
    See the third sentence of the first reply.
     
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