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Math Help - System in R2

  1. #1
    Super Member dhiab's Avatar
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    System in R2

    Find all real numbers a,b solutions for system :

    \left\{ \begin{array}{l}<br />
a^2 + 4b^2 = 1 \\ <br />
4a^2 + a + 2\sqrt 3 b = 2 \\ <br />
\end{array} \right.<br />
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  2. #2
    MHF Contributor
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    Quote Originally Posted by dhiab View Post
    Find all real numbers a,b solutions for system :

    \left\{ \begin{array}{l}<br />
a^2 + 4b^2 = 1 \\ <br />
4a^2 + a + 2\sqrt 3 b = 2 \\ <br />
\end{array} \right.<br />
    Hi

    From the first equation we can let a = \cos \theta and b = \frac12 \:\sin \theta

    The second equation gives

    4 \cos^2 \theta + \cos \theta + \sqrt3 \:\sin \theta = 2

    4 \cos^2 \theta + 2\:\left(\frac12\:\cos \theta + \frac{\sqrt3}{2} \:\sin \theta \right) = 2

    4 \cos^2 \theta + 2\:\cos \left(\theta - \frac{\pi}{3}\right) = 2

    2 \cos^2 \theta - 1 = -\:\cos \left(\theta - \frac{\pi}{3}\right)

    \cos(2 \theta) = \cos \left(\theta - \frac{\pi}{3} + \pi \right)

    Therefore
    2 \theta = \theta + \frac{2\pi}{3} + 2k \pi
    or
    2 \theta = -\theta - \frac{2\pi}{3} + 2k \pi

    \theta = \frac{2\pi}{3} + 2k \pi or \theta = - \frac{2\pi}{9} + 2k \frac{\pi}{3}

    Then
    (a,b) = \left(-\frac12,\frac{\sqrt{3}}{4}\right)

    (a,b) = \left(\cos\frac{-2\pi}{9},\frac12\:\sin\frac{-2\pi}{9}\right)

    (a,b) = \left(\cos\frac{4\pi}{9},\frac12\:\sin\frac{4\pi}{  9}\right)

    (a,b) = \left(\cos\frac{10\pi}{9},\frac12\:\sin\frac{10\pi  }{9}\right)
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