One Equation, One Variable!

The equation can be written as:

$x\sqrt{6} \left( \sqrt{3 - 2x + \sqrt{5(1-x^2)} } \right) = 2 -3x$

This equation for $0 < x < 2/3$ is equivalent to:

$6x^2 \left( 3 - 2x + \sqrt{5(1-x^2)} \right) = (2 -3x)^2$

Since $(3-2x)^2 - (2-3x)^2 = 5(1-x^2)$ , we can write the above equation as:

$6x^2 = \dfrac{(2 -3x)^2 (3 - 2x - \sqrt{5(1-x^2)}) }{(3-2x)^2 - 5(1-x^2)}$

or $6x^2 + 2x - 3 = - \sqrt{5(1-x^2)}$

Setting $\alpha = \arccos(2/3)$ and $x = \cos(\theta)$ , $\theta \in (\alpha, \pi /2)$ we get:

$-\sqrt{5}\sin(\theta) = 6\cos^2 (\theta) + 2\cos(\theta) - 3$

$\Rightarrow \dfrac{-\sqrt{5}\sin(\theta) - 2\cos(\theta)}{3} = 2\cos^2(\theta) - 1 = \cos(2\theta)$

$\Rightarrow \cos(\theta - \alpha) = \cos(\pi - 2\theta)$

Hence $\theta = \dfrac{\pi + \alpha}{3}, x = \cos \left( \dfrac{\pi + \arccos (2/3) }{3} \right)$ .