Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ where $x,y\in [0,\pi]$.
I am trying to  solve this but I am stuck. I know that $x=y=\pi/3$ is a solution but how do I show this is the only one? I think there are no others! Hints would be appreciated 

\begin{align}
   \cos x +\cos y - \cos(x+y) &= \frac 32 \\
   \cos x + \cos y - \cos x \cos y + \sin x \sin y 
      &= \frac 32 \\
   (1 - \cos y)\cos x + \sin y \sin x &= \frac 32 - \cos y \\
\end{align}
Note that
$$\sqrt{(1-\cos y)^2 + (\sin y)^2}  = \sqrt{1 - 2\cos y + \cos^2y + \sin^2 y} 
     = \sqrt{2(1 - \cos y)}
     = 2\sin \dfrac y2$$
$$\dfrac{1-\cos y}{2 \sin \dfrac y2} = \sin \dfrac y2$$
$$ \dfrac{\sin y}{2 \sin \dfrac y2}
        = \cos \dfrac y2$$
So
\begin{align}
   \sin \dfrac y2\cos x + \cos \dfrac y2 \sin x 
      &= \frac{\frac 32 - \cos y}{2 \sin \dfrac y2} \\
   \sin \left(x + \dfrac y2 \right)
      &= \frac{3 - 2\cos y}{4 \sin \dfrac y2} \\
   \sin \left(x + \dfrac y2 \right)
      &= \frac{1 + 4\sin^2 \dfrac y2}{4 \sin \dfrac y2} \\
   \sin \left(x + \dfrac y2 \right)
      &= \sin \dfrac y2 + \frac 14 \csc \dfrac y2\\
\end{align}
We can show that, for $y \in [0, \pi]$, the minimum acceptable value of 
$\sin \dfrac y2 + \frac 14 \csc \dfrac y2$ is $1$. The rest is pretty straight forward.
