Answer
$2\sin^{-1} \dfrac{x}{2} - \dfrac{2x \sqrt{4-x^2}}{4-2x^2}$
Work Step by Step
$\sin{\theta} =\dfrac{x}{2}$
$\cos{\theta} =\sqrt{1-\sin^2{\theta}} = \dfrac{1}{2} \sqrt{4-x^2}$
$\tan{\theta} =\dfrac{\sin{\theta}}{\cos{\theta}} = \dfrac{x}{\sqrt{4-x^2}}$
$\tan{2\theta} =\dfrac{2\theta}{1-\tan^2{\theta}} = \dfrac{2x \sqrt{4-x^2}}{4-2x^2}$
$2\theta - \tan{2\theta} = 2\sin^{-1} \dfrac{x}{2} - \dfrac{2x \sqrt{4-x^2}}{4-2x^2}$