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Thread: trig equation solve for theta

  1. #1
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    trig equation solve for theta

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

    for values of $\displaystyle \theta$ 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 $\displaystyle cos3\theta+cos\theta=0$

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

    convert $\displaystyle \sin\theta$ in terms of $\displaystyle cos\theta$

    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 $\displaystyle cos3\theta+cos\theta=0$

    for values of $\displaystyle \theta$ 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
    $\displaystyle cos3\theta = cos^{3}\theta - 3cos\theta sin^{2}\theta$

    convert $\displaystyle \sin\theta$ in terms of $\displaystyle cos\theta$

    can you solve now?
    would it be like this

    $\displaystyle cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0$ 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
    $\displaystyle \cos A = \cos B$
    has the general solution
    $\displaystyle A = 2n\pi \pm B$
    So, with the equation
    $\displaystyle 3\theta = 2n\pi \pm(\pi-\theta)$
    take the $\displaystyle +$ sign and $\displaystyle -$ sign one at a time; solve for $\displaystyle \theta$, and then plug in various values of $\displaystyle n\; (0, \pm1,\pm2,...)$ until you have all the values of $\displaystyle \theta$ 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.

    $\displaystyle 3\theta={\pi}-\theta$

    $\displaystyle 4\theta={\pi}$

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

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

    there are three more solutions until the solutions repeat themselves.


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

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

    $\displaystyle 2\theta=2n{\pi}-{\pi}$
    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

    $\displaystyle cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0$ 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 $\displaystyle sin^2(x)+ cos^2(x)= 1$, $\displaystyle sin^2(x)= 1- cos^2(x)$ so this becomes
    $\displaystyle cos^3(\theta)- 3cos(\theta)(1- cos^2(\theta)+ cos(\theta)= 0$

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

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

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

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

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