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Homework Help: Find all solutions of trig

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Find all solutions:

    [1] cos(2x) = 0
    [2] tan(5x) = 1


    2. Relevant equations

    3. The attempt at a solution

    For [1], this is my attempt:

    let u = 2x, du = 2 dx, so 1/2 du = x
    let me define a range, 0<= x <= pi
    so 0<= 1/2du <= pi
    and solve for du, which gives me 0 <= du <= 2pi

    cos(u) = 0 when u is pi/2, 3pi/2, and etc.
    cos(du) = 0, when pi/2, 3pi/2 and etc all divide by 2x
    this gives me pi/4, 3pi/4, and etc.
    it comes out to be n*pi/4, where n = odd


    for [2], i used the same approach

    tan(5x) = 1
    let u = 5x, 1/5du = dx,
    i define range of tan this 0 <= x < pi/2 (not including pi/2)
    so that replace x by 1/5 du and solve for du, i get new range 0<= du < 5pi/2

    for tan(u) = 1, i have possible u to be pi/4, 5pi/4
    and again, divide these by 1/5, i get tan(du) = 1 when pi/20, 5pi/20

    so i thought the answer would be n* pi/20
    now before i checked the answer, i asked myself, what about those limitation?
    tan has certain limitations in range, so it can't be just n * pi/20
    but i didn't figure it out

    when i checked the answer, it was pi/20 + n*pi/5
    so how did he get the last addition?

    Thanks
     
  2. jcsd
  3. Apr 6, 2010 #2

    Mark44

    Staff: Mentor

    That's domain, and isn't relevant to this problem. You're not asked to find an inverse function, so the domain is all real numbers.
    This is wrong. There's no calculus involved here; you're just asked to find the solutions of a trig equation.

    You have cos(2x) = 0. This means that 2x = p/2, 3pi/2, 5pi/2, ..., -pi/2, -3pi/2, -5pi/2, ...
    Yes. And etc. includes the negative values I showed above.
    You aren't integrating, so there's no need for du.
     
  4. Apr 6, 2010 #3
    Hi. thank you for your help.
    It seems like I have to find the trend, which sometime is not very clear.
    But I do understand why now.

    Thanks.
     
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