What is the value of x in the following problem?

Let $\theta = \sin^{-1} \left(\dfrac {3\sin 2\alpha}{5+4\cos 2\alpha}\right)$ . Then $\tan^{-1} x = \dfrac \theta 2 \implies x = \tan \dfrac \theta 2$ . Now

$\begin{aligned} \sin \theta & = \frac {3\sin 2\alpha}{5+4\cos 2\alpha} \\ \color{#3D99F6} \frac {2\tan \frac \theta 2}{1+ \tan^2 \frac \theta 2} & = \color{#D61F06} \frac {6\sin \alpha \cos \alpha}{5+8\cos^2 \alpha - 4} & \small \color{#3D99F6} \text{Note that }x = \tan \frac \theta 2 \\ \color{#3D99F6} \frac {2x}{1+x^2} & = \color{#D61F06} \frac {6\tan \alpha}{\sec^2 \alpha+8} & \small \color{#D61F06} \text{Divide up and down by }\cos^2 \alpha \\ & = \color{#D61F06} \frac {6\tan \alpha}{9 + \tan^2 \alpha} & \small \color{#D61F06} \text{Divide up and down by }9 \\ & = \color{#D61F06} \frac {\frac 23 \tan \alpha}{1 + \frac 19 \tan^2 \alpha} \\ \implies x & = \boxed{\frac 13 \tan \alpha} \end{aligned}$