<s>[INST] 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 
 [/INST] For $x,y \in [0,\pi]$
\begin{align}
  \cos x &= u \\
  \cos y &= v \\
  \sin x &= \sqrt{1-u^2} \\
  & \ge 0 \\
  \sin y &=\sqrt{1-v^2} \\
  & \ge 0 \\
  u+v-uv+\sqrt{1-u^2} \sqrt{1-v^2} &= \frac{3}{2} \\
  \sqrt{1-u^2} \sqrt{1-v^2} &= \frac{3}{2}+uv-u-v
\end{align}
By AM $\ge$ GM,
$$\frac{1-u^2+1-v^2}{2} \ge \frac{3}{2}+uv-u-v \tag{1}$$
equality holds if $1-u^2=1-v^2$ and rearranging $(1)$ gives
$$0 \ge \frac{(u+v-1)^2}{2} \tag{2}$$
But $\dfrac{(u+v-1)^2}{2}$ is non-negative which forces the equality to hold.
Therefore,
$$u+v=1 \land u^2=v^2$$
$$u+v=1 \land (u+v)(u-v)=0$$
$$u=v=\frac{1}{2} \implies x=y=\frac{\pi}{3}$$

Further points to be noticed
Considering the quadratic in $u$ of 
$$(1-u^2)(1-v^2)=\left( \frac{3}{2}+uv-u-v \right)^2$$
Discriminant $$\Delta_u =4v^4-4v^3-3v^2+4v-1=(v^2-1)(2v-1)^2$$
  
  
*
  
*$v^2>1 \implies \Delta_u>0$ but $v$ is beyond the range of cosine
  
*$v=1 \implies \Delta_u=0$ but gives $0=\frac{1}{4}$, so no solutions
  
*$v=-1 \implies \Delta_u=0$ but gives $u=\frac{5}{4}$ which is beyond the range of cosine
  
*$v=\frac{1}{2} \implies \Delta_u=0$ that gives $u=\frac{1}{2}$
  
*$\Delta_u<0$ elsewhere

</s>