$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$
$\begin{array}{r}
\therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-\left(2 \cos ^2\left(\frac{x+y}{2}\right)-1\right)=\frac{3}{2} \\
\quad \cdots\left[\begin{array}{l}
\because \cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) \text { and } \\
\cos \theta=2 \cos ^2\left(\frac{\theta}{2}\right)-1
\end{array}\right]
\end{array}$
$\begin{aligned} & \therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \cos ^2\left(\frac{x+y}{2}\right)=\frac{3}{2}-1 \\ & \therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \cos ^2\left(\frac{x+y}{2}\right)=\frac{1}{2} \\ & \therefore \quad 4 \cos ^2\left(\frac{x+y}{2}\right)-4 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)+1=0\end{aligned}$
Substituting $\cos \left(\frac{x+y}{2}\right)=\mathrm{t}$, we get
$4 \mathrm{t}^2-4 \mathrm{t} \cos \left(\frac{x-y}{2}\right)+1=0$
As $t$ is real, we get $b^2-4 a c \geq 0$
$\begin{aligned}
& \Rightarrow\left[-4 \cos \left(\frac{x-y}{2}\right)\right]^2-4 \times 4 \times 1 \geq 0 \\
& \Rightarrow 16 \cos ^2\left(\frac{x-y}{2}\right)-16 \geq 0 \\
& \Rightarrow \cos ^2\left(\frac{x-y}{2}\right) \geq 1 \\
& \Rightarrow \cos ^2\left(\frac{x-y}{2}\right)=1
\end{aligned}$
$\ldots[\because-1 \leq \cos \theta \leq 1$, for all values of $\theta]$
$\begin{aligned}
& \Rightarrow \frac{x-y}{2}=0 \\
& \Rightarrow x=y
\end{aligned}$