How to prove whether the equation set has a unique solution? \begin{eqnarray}
\begin{cases}
\sin A \sin C-(\sin B)^2=0
\cr AC-B^2=0
\cr A+B+C-\pi=0
\cr A>0,B>0,C>0
\end{cases} \end{eqnarray} 
How to prove whether the equation set has a unique solution or not ?
 A: The third and fourth conditions tell me you are investigating a triangle. So fixing points $A$ and $B$ of a triangle $ABC$, I plotted the first two conditions depending on where $C$ might lie:

The color code is as follows:
\begin{alignat*}{2}
\text{Blue:}&\qquad& \sin A\sin C-\sin^2B &> 0 &\qquad AC-B^2 &> 0 \\
\text{Magenta:}&& \sin A\sin C-\sin^2B &> 0 &\qquad AC-B^2 &< 0 \\
\text{Yellow:}&& \sin A\sin C-\sin^2B &< 0 &\qquad AC-B^2 &< 0
\end{alignat*}
So the inner boundary line between blue and magenta is where the second equation is satisfied, while the outer boundary line between magenta and yellow is where the first equation is satisfied. Both equations can only be satisfied simultaneously if the two curves touch, which they only do in the labeled point $C$, the corner of a regular triangle.
Now that I think about it, that plot is needlessly breaking the symmetry of the situation. I should fix $A$ and $C$ and plot depending on the position of $B$:

Of course, such a plot is not a proof, but it might help others in finding a proof. And for me personally it is enough to convince me that the regular triangle is indeed the only solution here.
For those interested, the plot displayed above was created with Cinderella, using its colorplot function.
A: Assume WLOG due to symmetry that $A\leq C$ and let us rule out $A<C$. So suppose $A<C$. Then by the second equation $B=\sqrt{AC}$ so then $A<B<C$.
If the first equation holds there is some constant $\gamma>0$ such that
$$
\gamma=\frac{\sin(B)}{\sin(A)}=\frac{\sin(C)}{\sin(B)}
$$
Similarly, if the second equation holds we have another constant $\lambda>0$ such that
$$
\lambda=\frac BA=\frac CB
$$
Since $\sin(x)$ is concave on the interval $[0,\pi]$ the line through $(0,0)$ and $(A,\sin(A))$ passes above the point $(B,\sin(B))$, so we have the relation of slopes
$$
\frac{\sin(B)}{B}<\frac{\sin(A)}{A}\implies \gamma<\lambda
$$
Combining these informations, we see that the three points
$$
\begin{align}
(A,\sin(A))&\\
(B,\sin(B))&=(\lambda A,\gamma\sin(A))\\
(C,\sin(C))&=(\lambda^2 A,\gamma^2\sin(A))
\end{align}
$$
must lie on the graph of the same power function $f(x)=a\cdot x^b$ for some $a>0$ and $0<b<1$. But this is impossible, since
$$
g(x)=\sin(x)-a\cdot x^b
$$
has derivative $g'(x)=\cos(x)-ab\cdot x^{b-1}$, and $x\mapsto ab\cdot x^{b-1}$ is positive and convex (since $b-1<0$) so it intersects the positive part of $x\mapsto \cos(x)$, which is concave, at most twice. Thus $g'(x)$ has at most two zeros and $g(x)$ has at most three zeros, one being at $x=0$. So $g(A)=g(B)=g(C)=0$ is impossible for any triple $0<A<B<C$.

So this implies that we can only hope to find a solution where $A=C$. Then quickly $A=B=C=\frac{\pi}{3}$ pops out as the only solution.
