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Math Help - cos(x -2pi/3) = 1/2

  1. #1
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    cos(x -2pi/3) = 1/2

    Hi,

    the question I have been working on is cos(x -2pi/3) = 1/2 -2pi<x<2pi

    I know from one of the special triangles that cos(pi/3) = 1/2 = cos(x-2pi/3)

    so cos(x-2pi/3) = cos(pi/3), cos(-pi/3), cos(5pi/3), cos(-5pi/3), cos(7pi/3), cos(-7pi/3)

    if this is right, should I just solve for each of them eg:

    if cos(x-2pi/3) = cos(pi/3) ==> x-2pi/3 = pi/3 ==> x = 2pi/3+ pi/3 ==> x = pi


    It would also be good if someone could let me know if I have missed any angles that give cos()= 1/2

    thanks
    Last edited by andy000; March 20th 2014 at 05:17 AM.
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  2. #2
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    Re: cos(x -2pi/3) = 1/2

    Quote Originally Posted by andy000 View Post
    Hi,

    the question I have been working on is cos(x -2pi/3) = 1/2 -2pi<x<2pi

    I know from one of the special triangles that cos(pi/3) = 1/2 = cos(x-2pi/3)

    so cos(x-2pi/3) = cos(pi/3), cos(-pi/3), cos(5pi/3), cos(-5pi/3), cos(7pi/3), cos(-7pi/3)

    if this is right, should I just solve for each of them eg:

    if cos(x-2pi/3) = cos(pi/3) ==> x-2pi/3 = pi/3 ==> x = 2pi/3+ pi/3 ==> x = pi


    It would also be good if someone could let me know if I have missed any angles that give cos()= 1/2

    thanks
    cos 60 = 1/2
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  3. #3
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    Re: cos(x -2pi/3) = 1/2

    Quote Originally Posted by bjhopper View Post
    cos 60 = 1/2
    That is not correct. $\cos 60^\circ = \dfrac{1}{2}$, but $\cos 60 \approx -0.95$.

    Your value of $\dfrac{7\pi}{3}$ implies $x = 3\pi > 2\pi$, so that $x$ value is not in the domain. (But $\cos(x-2\pi/3) = \cos(-7\pi/3)$ does give an $x$-value in the domain).
    Last edited by SlipEternal; March 20th 2014 at 07:38 AM.
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  4. #4
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    Re: cos(x -2pi/3) = 1/2

    Hi,

    Thanks for the reply,

    so I guess I should have four values for x (discarding cos(7pi/3) and cos(5pi/3)?
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  5. #5
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    Re: cos(x -2pi/3) = 1/2

    Quote Originally Posted by andy000 View Post
    Hi,

    Thanks for the reply,

    so I guess I should have four values for x (discarding cos(7pi/3) and cos(5pi/3)?
    That is correct.
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  6. #6
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    Re: cos(x -2pi/3) = 1/2

    Thanks again =)
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  7. #7
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    Re: cos(x -2pi/3) = 1/2

    Hello, andy000!

    \text{Solve: }\:\cos\left(x -\tfrac{2\pi}{3}\right) \:=\:\tfrac{1}{2}\;\text{ for }-2\pi\,<\,x\,<\,2\pi

    We have: . \cos\left(x - \tfrac{2\pi}{3}\right) \:=\:\tfrac{1}{2}

    . . x - \tfrac{2\pi}{3} \;=\;\pm\tfrac{\pi}{3} + 2\pi n

    . . x - \tfrac{2\pi}{3} \;=\;\begin{Bmatrix}\text{-}\frac{7\pi}{3} \\ \text{-}\frac{5\pi}{3} \\ \text{-}\frac{\pi}{3} \\ \frac{\pi}{3} \end{Bmatrix}

    . . x \;=\;\begin{Bmatrix}\text{-}\frac{5\pi}{3} \\ \text{-}\pi \\ \frac{\pi}{3} \\ \pi \end{Bmatrix}
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  8. #8
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    Re: cos(x -2pi/3) = 1/2

    Hi Soroban,

    Thanks for your reply. Can you explain how you get the equation x-2pi/3 = +/- pi/3 + 2pin?

    I know this to be true but only from drawing out the unit circle and looking at it and the angles to equal pi/3. Is there some understanding I could gain to help me do this for other problems in the future?

    Thanks
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