Find value of $\cot(\theta-\alpha)$ Let $\theta$ and $\alpha$ be the solutions to $$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=\dfrac{3\cos2x+6}{10}$$
find the value of $\cot(\theta-\alpha)$, where $\dfrac{\pi}{2}<\theta<\alpha<\dfrac{3\pi}{2}$
My attempt :
Let $A=\dfrac{x}{2}$, $$\dfrac{\sin^2A}{1-\cot A}+\dfrac{\cos^2A}{1-\tan A}=\dfrac{3\cos 4A+6}{10}$$
$$\dfrac{\sin A\cos A-\sin^4A-\cos^4A}{2\sin A\cos A-\cos^2A-\sin^2A}=\dfrac{12\cos^2A-3}{10}$$
$$\dfrac{\sin A\cos A+\cos^2A-\sin^2A}{2\sin A\cos A-1}=\dfrac{12\cos^3A-3}{10}$$
Are there any better ways for approaching this problem? or I've come the right way.

Note that 
$$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=
\frac{\sin^3\frac x2 -\cos^3\frac x2}{\sin\frac x2 - \cos\frac x2}
=1+\frac12\sin x
$$
which, along with $\frac{3\cos2x+6}{10}= \frac{9-6\sin^2 x}{10} $, leads to
$$6\sin^2 x +5\sin x +1=0$$
Then, $\sin\alpha =-\frac12$, $ \sin \theta =-\frac13$ and correspondingly $\tan\alpha = \frac1{\sqrt3}$, $\tan\theta = \frac1{2\sqrt2}$
$$\cot(\theta- \alpha) = \frac{1+\tan \theta \tan\alpha }{\tan \theta - \tan \alpha} 
=-\frac{8\sqrt2+9\sqrt3}{5}
$$
