How to solve this complicated trigonometric equation for the solution of this triangle. 
In $\triangle ABC$, if $AB=AC$ and internal bisector of angle $B$ meets $AC$ at $D$ such that $BD+AD=BC=4$, then what is $R$ (circumradius)?


My Approach:- I first drew the diagram and considered $\angle ABD=\angle DBC=\theta$ and as $AB=AC$, $\angle C=2\theta$.  Therefore $\angle A=180-4\theta$. Also as $AE$ is the angular bisector and $AB=AC$, then $BE=EC=2.$ Now applying the sine theorem to $\triangle ADB$ and  $ \triangle BDC$ gives $$\frac{BD}{\sin {(180-4\theta)}}=\frac{AD}{\sin \theta}$$
$$\frac{BC}{\sin(180-3\theta)}=\frac{BD}{\sin 2\theta}$$
Now we know that $BC=4$ and then solving both the equations by substituting in $BD+AD=4$, we get $$\sin 2\theta .\sin4\theta+\sin2\theta.\sin\theta=\sin3\theta.\sin4\theta$$
$$\sin4\theta+\sin\theta=\frac{\sin3\theta.\sin4\theta}{\sin2\theta}$$
Now I have no clue on how to proceed further from here. Though I tried solving the whole equation into one variable ($\sin\theta$), but it's getting very troublesome as power of $4$ occurs. Can anyone please help further or else if there is any alternative method to solving this problem more efficiently or quickly?
Thank You

The values are such that equations do not simplify. Nonetheless, here is an alternate approach.
Say $AB = AC = x$ and we know $BC = 4$,
By angle bisector theorem,
$\cfrac{4}{x} = \cfrac{x-AD}{AD} \implies AD = \cfrac{x^2}{4+x}$
Now by angle bisector length formula,
$BD^2 = \cfrac{4x}{(4+x)^2} [(4+x)^2 - x^2] = \cfrac{32x(x+2)}{(4+x)^2}$
Now, $BD + AD = 4 = \cfrac{x^2}{4+x} + \cfrac{\sqrt{32x(x+2)}}{4+x}$
$16+4x-x^2 = \sqrt{32x(x+2)}$
Solving using WolframAlpha, the only valid solution is $x \approx 2.61$
Now to find circumradius, use $R = \cfrac{abc}{4 \triangle} = \cfrac{4 x^2}{8 \sqrt{x^2 - 4}}$
