If $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$, then is it unique and its area? Given that $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$. Then would the $\triangle ABC$ be unique? If so how easily we may find its area or otherwise the maximum possible area? 
Any help is appreciated :)
 A: By law of sines, we have
$$\sin A=\frac{a}{2R}\Rightarrow 1-\left(\frac{1}{2R}\right)^2=\frac{a^2}{4R^2}\Rightarrow a=\sqrt{4R^2-1}.$$
Also, we have
$$b=\sqrt{4R^2-4},\ \ \ \ c=\sqrt{4R^2-9}.$$
Then, the area $S$ is
$$S=\frac 12bc\sin A=\frac{\sqrt{(4R^2-1)(4R^2-4)(4R^2-9)}}{4R}.$$
A: Since
$$
\cos(B)=2\cos(A)\tag{1}
$$
and
$$
\cos(C)=3\cos(A)\tag{2}
$$
and furthermore, since $A+B+C=\pi$,
$$
\cos(C)=-\cos(A+B)\tag{3}
$$
Using $(1)$ and $(2)$ in $(3)$ says
$$
3\cos(A)=\sin(A)\sin(B)-2\cos^2(A)\tag{4}
$$
Moving $2\cos^2(A)$ to the left and squaring yields
$$
\begin{align}
4\cos^4(A)+12\cos^3(A)+9\cos^2(A)&=(1-\cos^2(A))(1-4\cos^2(A))\\
12\cos^3(A)+14\cos^2(A)-1&=0\\
\cos(A)&=0.24312617957188074493\tag{5}
\end{align}
$$
Since $R=\dfrac1{2\cos(A)}$ and the area is $2R^2\sin(A)\sin(B)\sin(C)$, we get
$$
\begin{align}
\text{Area}
&=\frac{\sqrt{1-\cos^2(A)}\sqrt{1-4\cos^2(A)}\sqrt{1-9\cos^2(A)}}{2\cos^2(A)}\\
&=4.90482198456130394784\tag{6}
\end{align}
$$
A: Note $A+B+C=\pi,\cos {(A+B+C)}=-1 \implies 1-\cos(A)\sin(B)\sin(C)-\sin(A)\cos(B)\sin(C)-\sin(A)\sin(B)\cos(C)+\cos(A)\cos(B)\cos(C)=0$
let $D=2R \implies D \ge 3 $, we have $\sin A= \sqrt{1-\cos^2 A}=\dfrac{\sqrt{D^2-1^2}}{D},\sin B=\dfrac{\sqrt{D^2-2^2}}{D},\sin C=\dfrac{\sqrt{D^2-3^2}}{D}$
$D^3+1*2*3=3\sqrt{D^2-1^2}\sqrt{D^2-2^2}+2\sqrt{D^2-1^2}\sqrt{D^2-3^2}+1*\sqrt{D^2-3^2}\sqrt{D^2-2^2} \iff (2D+12)*\sqrt{(D^2-9)}\sqrt{(D^2-4)}=D^4-12D^2+12D+72 \iff D^3-14D-12=0 \iff D=4.1131$
which means the triangle is unique. 
