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Find All Real Solutions (problem)

  1. Jul 14, 2003 #1


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    Problem is:

    2 cos 3x - 1 = 0

    Here's how I've done it so far:

    2 cos 3x = 1

    cos 3x = 1/2

    Now I'm stuck. Where do I go from here? Do I divide the 3 out making it cos x = 1/6? I know how to find all the real solutions once I find what cos x is. I'm just having trouble figure out exactly how to get cos x in that type of problem.

    Any help is grealy appreciated.
    Last edited: Jul 14, 2003
  2. jcsd
  3. Jul 14, 2003 #2
    I think you meant cos x = 1/6. No, we can't take the 3 out.

    let y = 3x
    cos y = 1/2
    y = [the general solution]
    3x = [the general solution]
    x = 1/3 * [the general solution]
  4. Jul 15, 2003 #3
    just let me make the [general solution] a little bit clearer.
    Now, we first define y=3x
    therefore :
    cos y = 1/2
    or :
    y = cos-1(1/2) + 2n[pi]
    Now, you might be thinking "Where did 2n[pi] come from ?"
    Well, you must remember that Cos() repeats itself after 2[pi], 4[pi], 6[pi], ... , or in other words after 2n[pi] (where n is an integer), so the value of cos(z) and cos(z+2n[pi]) is the same.
    Now, from the last equation, you must either use a calculator to figure out cos-1(1/2), or try to remember if the number looks familiar.
    The number looks familiar to me, i know that Cos([pi]/3) = 1/2 , or
    Cos-1(1/2) = [pi]/3
    So, now, you can rewrite your equation, and solve it.
  5. Jul 15, 2003 #4
    I think it would be quite useful if you can draw a y = cos x graph.

    Anyway, you just need to find all the solutions where cos x = 1/2. Then divide x by a factor of 3.
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