Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find All Real Solutions (problem)

  1. Jul 14, 2003 #1


    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook