System in R2

**dhiab** · August 15th 2009, 09:54 AM

Find all real numbers a,b solutions for system :

$\left\{ \begin{array}{l} a^2 + 4b^2 = 1 \\ 4a^2 + a + 2\sqrt 3 b = 2 \\ \end{array} \right. $

**running-gag** · August 15th 2009, 12:10 PM

Originally Posted by dhiab

Find all real numbers a,b solutions for system :

$\left\{ \begin{array}{l} a^2 + 4b^2 = 1 \\ 4a^2 + a + 2\sqrt 3 b = 2 \\ \end{array} \right. $

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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