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Math Help - problems with cosine equation

  1. #1
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    Exclamation problems with cosine equation

    i have 2cos(2x)-2cos(x)=0 and im trying to find the values of x. i already figured out that they are +/- 2pi/3 but i dont know how to work that out showing work. all i have so far is cos2x=cosx but i dont know what to do with that. any help at all would be appreciated since i do not know how to approach this, thank u!
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    Quote Originally Posted by airportman92 View Post
    i have 2cos(2x)-2cos(x)=0 and im trying to find the values of x. i already figured out that they are +/- 2pi/3 but i dont know how to work that out showing work. all i have so far is cos2x=cosx but i dont know what to do with that. any help at all would be appreciated since i do not know how to approach this, thank u!
    nevermind, i got it i think!
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by airportman92 View Post
    i have 2cos(2x)-2cos(x)=0 and im trying to find the values of x. i already figured out that they are +/- 2pi/3 but i dont know how to work that out showing work. all i have so far is cos2x=cosx but i dont know what to do with that. any help at all would be appreciated since i do not know how to approach this, thank u!

    Note that \cos\!\left(2x\right)=2\cos^2x-1.

    So we get 2\cos^2x-1-\cos x=0

    Note that this is quadratic in \cos x. So letting u=\cos x, we get 2u^2-u-1=0.

    I leave it for you to solve for u (you'll get 2 different values). Then set \cos x equal to the values and solve for x. Does this make sense?
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  4. #4
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    Smile

    .....sorry.....
    wrong post
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  5. #5
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    Hi there airportman92

    Your equation

     2\cos(2x)-2\cos(x)=0

    needs a substitution \cos(2x)=2\cos^2(x)-1

    Therefore you have  2(2\cos^2(x)-1)-2\cos(x)=0

    Now expanding gives you

     4\cos^2(x)-2-2\cos(x)=0

    dividing through by 2 gives

     2\cos^2(x)-1-\cos(x)=0

    and re arranging

     2\cos^2(x)-\cos(x)-1=0

    now making a = \cos(x)

     2a^2-a-1=0

    finally solve the quadratic...
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  6. #6
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    ooo ok thanks!

    what about its derivative which i got to be -4sin2x+2sinx.
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  7. #7
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    correct!

     (2\cos(2x)-2\cos(x))'= -4\sin(2x)+2\sin(x)
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  8. #8
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    Quote Originally Posted by pickslides View Post
    correct!

     (2\cos(2x)-2\cos(x))'= -4\sin(2x)+2\sin(x)
    but the x-values are impossible to solve for. i know its supposed to be cos^-1(.25) but theres no way to show that
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  9. #9
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    Quote Originally Posted by airportman92 View Post
    but the x-values are impossible to solve for. i know its supposed to be cos^-1(.25) but theres no way to show that

    Are you talking about the original question or something to do with the derivative?
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  10. #10
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    Quote Originally Posted by pickslides View Post
    Are you talking about the original question or something to do with the derivative?

    well this is already the derivative of somethin, so i just neeed the x value which i thought i got
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  11. #11
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    Quote Originally Posted by airportman92 View Post
    well this is already the derivative of somethin, so i just neeed the x value which i thought i got

    Did you want to solve for x with respect to the original equation or the derivative? 2 different things!
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