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Trigonomic Equation

  1. Feb 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve the following trigonomic equation for 0<=x360<= :
    5cos2x + cosx + 2 = 0

    2. Relevant equations

    Not sure what goes here but I've been trying to substitute these -
    cos2x = cos^2x - sin^2x
    cos2x = 2cosx^2 - 1 - I think this is the correct one to use.
    cos2x = 1 - sin^2x

    3. The attempt at a solution

    5cos2x + cosx + 2 = 0
    5(2cos^2x-1) + cosx + 2 = 0
    10cos^2x + cosx - 3 =0
    cosx (10cosx +1) - 3 = 0
    ? I get stuck here

    Am I going about this the right way, or am I completely lost? I think I'm doing it right, I'm just not sure how to continue.

    Any help appreciated.
  2. jcsd
  3. Feb 7, 2007 #2


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    From here: 10cos^2x + cosx - 3 =0, make the substitution cosx=y, and solve the quadratic equation for y. Then, say you have solutions y=a,b, you can then solve the equations a=cosx, b=cosx to obtain all values of x.
  4. Feb 7, 2007 #3
    Hi. Thanks for the reply.

    I've never been taught how to do that, so is there another way of doing it?

    Thanks again,
  5. Feb 7, 2007 #4


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    You have a quadratic in cosx. Now, the usual ways to solve the equation are to either use the quadratic equation, or to spot some factors and factorise the expression. I used the substitution y=cosx to simplify the algebra throughout the calculation of the roots of the quadratic equation.

    Perhaps your teacher hasn't explicitly taught you this, since there isn't really anything to be taught here! He expects you to be able to solve a quadratic equation, and here you have one in cosx.
  6. Feb 7, 2007 #5
    Thank you!

    I just figured out what you were saying. I didn't see it as a quadratic equation at all - oh dear (exam in 4 day, haha).

    Figured it out, now just need to calculate the actual angles invovled. Thanks again.

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