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Finding where x = 0

  1. May 26, 2015 #1
    It's been a long time since I had to try and figure this stuff out, so a little refresher would be helpful.

    I'm trying to find where x = 0 in this equation: sin((x-10)^2)

    I forget how this works. Can I get a little help?

    Thank you!
     
  2. jcsd
  3. May 26, 2015 #2

    SteamKing

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    You have an expression, not an equation.

    Are you instead trying to find out the values of x which make sin ((x-10)2) = 0 ?
     
  4. May 26, 2015 #3
    Right, yes! That!
     
  5. May 26, 2015 #4

    phion

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    I believe you mean to find where [itex]sin((x-10)^2)) = 0[/itex],where [itex]x^2-20x+100=n\pi[/itex].
     
  6. May 26, 2015 #5

    Mark44

    Staff: Mentor

    There's no advantage in expanding (x - 10)2 in the above. Just solve the equation ##(x - 10)^2 = n\pi## for x.
     
  7. May 26, 2015 #6
    So, x = (√(nπ) + 10)? So then where does arcsin come in?
     
  8. May 26, 2015 #7

    SteamKing

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    sin ((x- 10)2) = 0

    arcsin [sin ((x-102)] = arcsin (0)

    (x - 10)2 = nπ, where n = 0, 1, 2, 3, ...

    x - 10 = √(nπ)

    x = √(nπ) + 10

    How about that?
     
  9. May 26, 2015 #8

    Mark44

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    Make that ##x - 10 = \pm\sqrt{n\pi}## and I'll be happy.
     
  10. May 26, 2015 #9
    aaaaaaah nice, it's kind of coming trickling back, now. Thank you, sirs!
     
  11. Jun 2, 2015 #10
    OK, I'm confused again. So...

    arcsin [sin ((x-102)] = arcsin (0) = (x - 10)2 = nπ ...?

    How does arcsin(0) equal anything but 0?
     
  12. Jun 2, 2015 #11

    Mark44

    Staff: Mentor

    It doesn't. In my opinion you'll just confuse yourself by invoking arcsin.

    You're trying to solve the equation ##\sin((x - 10)^2) = 0##. The easiest way to solve this is to recognize that sin(A) = 0 when A is any integer multiple of ##\pi##, so the solution of this equation is ##A = k\pi##, where k is any integer. Look at a graph of y = sin(x) to see this.

    Going back to your equation, you have ##\sin((x - 10)^2) = 0##, so it must be that ##(x - 10)^2 = k\pi##, with k again being any integer. Can you solve this equation for x now?
     
  13. Jun 3, 2015 #12
    OK, that helps. I've actually been working with a much more complex expression that turns out I would need to use imaginary numbers to fully plot, so that hasn't helped. And also I've been mapping a series of formulas and writing checks into excel, which doesn't throw zero on sin(pi), so I'm getting some red herrings. But once I've accounted for these things and done them manually or with simpler functions I've had more luck.

    Many thanks!
     
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