If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$. 
If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.

HINT: The answer is $-2\pm \sqrt5$.
What I have tried so far:
$$\tan{\frac{x}{2}} = \frac{1}{\sin x}-\sin x$$
$$\tan{\frac{x}{2}} = \frac{1-\sin^2 x}{\sin x}$$
$$\tan{\frac{x}{2}} = \frac{\cos^2 x}{\sin x}$$
I don't know how to solve this problem. Pls help. Thank you :)

Notice that
$$
\cos(x) = \frac{1 -\tan^2 (x/2)}{1+\tan^2(x/2)}
$$
And that
$$
\tan^2(x/2) = \frac{\cos^4(x)}{1-\cos^2(x)}
$$
Let $t = \cos(x)$. Plug the 2nd equation into the first one and after some algebra, we get
$$
(1-t)(1-t^2-t^4) = 0
$$
