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Math Help - trig equation solve for theta

  1. #1
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    trig equation solve for theta

    hello how would i solve for cos3\theta+cos\theta=0

    for values of \theta between 0 and 360 degrees
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by sigma1 View Post
    hello how would i solve for cos3\theta+cos\theta=0

    for values of \theta between 0 and 360 degrees
    cos3\theta = cos^{3}\theta - 3cos\theta sin^{2}\theta

    convert \sin\theta in terms of cos\theta

    can you solve now?
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  3. #3
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    Quote Originally Posted by sigma1 View Post
    hello how would i solve for cos3\theta+cos\theta=0

    for values of \theta between 0 and 360 degrees
    You can rewrite the equation as
    cos(3θ) = -cos(θ) = cos(π - θ)
    Therefore the general solution is
    3θ = 2nπ + or - (π - θ)
    Now solve for θ.
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  4. #4
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    Quote Originally Posted by harish21 View Post
    cos3\theta = cos^{3}\theta - 3cos\theta sin^{2}\theta

    convert \sin\theta in terms of cos\theta

    can you solve now?
    would it be like this

    cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0 if so where do i go from here? can u help me solve this?
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  5. #5
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    Hello sigma1

    sa-ri-ga-ma's solution is the easier method. He has used the fact that
    \cos A = \cos B
    has the general solution
    A = 2n\pi \pm B
    So, with the equation
    3\theta = 2n\pi \pm(\pi-\theta)
    take the + sign and - sign one at a time; solve for \theta, and then plug in various values of n\; (0, \pm1,\pm2,...) until you have all the values of \theta in the range you want.

    Grandad
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  6. #6
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    Some graphs for sa-ri-ga-ma's solution.

    3\theta={\pi}-\theta

    4\theta={\pi}

    3\theta=2n{\pi}+({\pi}-\theta)

    4\theta=2n{\pi}+{\pi}

    there are three more solutions until the solutions repeat themselves.


    3\theta=2n{\pi}-({\pi}-\theta)

    3\theta=2n{\pi}-{\pi}+\theta

    2\theta=2n{\pi}-{\pi}
    Attached Thumbnails Attached Thumbnails trig equation solve for theta-cosx-cos3x.jpg   trig equation solve for theta-cos3x-cosx.jpg  
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  7. #7
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    Quote Originally Posted by sigma1 View Post
    would it be like this

    cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0 if so where do i go from here? can u help me solve this?
    harish21 said "convert sine in terms of cosine" and you have not done that yet. Since sin^2(x)+ cos^2(x)= 1, sin^2(x)= 1- cos^2(x) so this becomes
    cos^3(\theta)- 3cos(\theta)(1- cos^2(\theta)+ cos(\theta)= 0

    cos^3(\theta)- 3 cos(\theta)+ 3 cos^3(\theta)+ cos(\theta)= 0

    4 cos^3(\theta)- 2 cos(\theta)= 0

    If you like, you can replace cos(\theta) with "x" so you have the equation
    4x^3- 2x= x(4x^2- 2)= 0

    Either x= 0 or 4x^2- 2= 0 which leads to x^2= \frac{1}{2} so that x= \pm\frac{1}{\sqrt{2}}= \pm\frac{\sqrt{2}}{2}.

    Now you need to solve y= cos(\theta)= 0, cos(\theta)= \frac{\sqrt{2}}{2}, and cos(\theta)= -\frac{\sqrt{2}}{2}.
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