<s>[INST] 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
 [/INST] Indeed you are proceeding correctly. The equation can be solved as follows:
$$\sin 2\theta \sin \theta=\sin 4\theta(\sin 3\theta-\sin 2\theta) {\tag 1}$$
Now $\sin 3\theta-\sin 2\theta=2\cos \frac {5\theta}{2} \sin \frac {\theta}{2}$.
Also $\sin \theta=2\cos \frac {\theta}{2} \sin \frac {\theta}{2}$, and $\sin 4\theta=2\sin 2\theta \cos 2\theta$.
So, $(1)$ simplifies to:
$$\cos \frac {\theta}{2}=2\cos \frac {5\theta}{2}\cos 2\theta {\tag 2}$$
Since $2\cos \frac {5\theta}{2} \cos 2\theta=\cos \frac {9\theta}{2}+\cos \frac {\theta}{2}$, we have, from $(2)$:
$$\cos \frac {9\theta}{2}=0$$
This means that $\frac {9\theta}{2}=\frac {\pi}{2}$, hence $\theta=\frac {\pi}{9}$.
This means that all angles of triangle are known, and we know $BC=4$. Thus using sine law, $$2R=\frac {BC}{\sin A}$$ is easy to calculate.
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