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18cos^2x+3cosx-1 = 0

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data


    it's a quadratic equation whose solutions are x= 1.4033+-2npi and x=1.9106+-2npi

    2. Relevant equations
    3. The attempt at a solution

    The answers to this problem are given as x= +-1.4033+-2npi and x=+-1.9106+-2npi

    Why are x's both positive and negative?

  2. jcsd
  3. Jan 26, 2012 #2
    Re: 18cos^2x+3cosx-1=0

    I think you're missing another answer i think

  4. Jan 26, 2012 #3
    Re: 18cos^2x+3cosx-1=0

    cos is an even function.
  5. Jan 26, 2012 #4

    D H

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    Re: 18cos^2x+3cosx-1=0

    That answer is there. It is the same as -1.9106±2npi.

    To solve this, first substitute u=cos(x). With this, 18u2+3u-1 is a quadratic with solutions u=cos(x)=1/6 and u=cos(x)=-1/3. Each of these solutions for cos(x) will generate two families of solutions for x because cos(x)=cos(-x).

    Note that you will also get four families of solutions for 18sin2x+3sin x + 1 = 0, but now the relevant symmetric is sin(pi-x)=sin(x) rather than cos(x)=cos(-x).
  6. Jan 26, 2012 #5
    Re: 18cos^2x+3cosx-1=0

    How are those solutions equal?
    Can I extend the identity cos(x)=cos(-x) to this problem below too?


    The answer given in my book: x=pi/12+-npi/2 and x=5pi/12+-npi/2

    Can I write: x=+-pi/12+-npi/2 and x=+-5pi/12+-npi/2?
  7. Jan 26, 2012 #6
    Re: 18cos^2x+3cosx-1=0

    like mindscrape said cos is an even function.
  8. Jan 26, 2012 #7

    D H

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    Re: 18cos^2x+3cosx-1=0

    Your solution is [itex]-1.9106\pm2n\pi, n\in\mathbb N[/itex], mtayab1994's is [itex]4.3725\pm2n\pi,n\in\mathbb N[/itex].

    First, get rid of the ± in front of the [itex]2n\pi[/itex]. It's not really needed. Just let n range over all the integers instead of over the non-negative integers. Next, change your n to m. No change here; that n or m is just a label. So now we have [itex]-1.9106+2m\pi, m\in\mathbb Z[/itex] versus mtayab1994's [itex]4.3725+2n\pi, n\in\mathbb Z[/itex]. Now set your m to n+1. Again there is no change to the solutions here; your n+1 is still ranging from -∞ to +∞. Your solution becomes [itex]-1.9106+2m\pi = -1.9106+2(n+1)\pi = (2\pi-1.9106) + 2n\pi = 4.3726 + 2n\pi[/itex]. That difference of 0.0001 is just rounding error. They are the same solution.

    Sure. [itex]\cos4x=1/2[/itex] means [itex]4x=\pm\pi/3+2n\pi[/itex], or [itex]x=\pm\pi/12+n\pi/2[/itex]. With a little effort you should be able to show that the [itex]x=-\pi/12+n\pi/2[/itex] family of solutions and the textbook's [itex]x=5\pi/12+n\pi/2[/itex] are one and the same.

    That is something you should not do. The reason is that your four families of solutions are really just two families.
  9. Jan 27, 2012 #8
    Re: 18cos^2x+3cosx-1=0

    Thank You. It's a bit over my head, but I'll try and work through it.

    Also appriciate all the other answers in this thread.
  10. Jan 27, 2012 #9


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    Re: 18cos^2x+3cosx-1=0

    If we have an answer of, say, 1+2k for k all integers, then this is the same as -1+2k, or 3+2k, or 1001+2k because all these answers are still the same set ...,-3, -1, 1, 3, 5,...
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