If \(\cos x+\cos y-\cos (x+y)=\frac{3}{2}\), then

\(\cos x+\cos y-\cos (x+y)=\frac{3}{2}\)
\(\begin{array}{r}
\therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-\left(2 \cos ^2\left(\frac{x+y}{2}\right)-1\right)=\frac{3}{2} \\
\quad \cdots\left[\begin{array}{l}
\because \cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) \text { and } \\
\cos \theta=2 \cos ^2\left(\frac{\theta}{2}\right)-1
\end{array}\right]
\end{array}\)
\(\therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \cos ^2\left(\frac{x+y}{2}\right)\)\(=\frac{3}{2}-1 \)
\(\begin{aligned} & \therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \cos ^2\left(\frac{x+y}{2}\right)=\frac{1}{2} \end{aligned}\)
\(\therefore \quad 4 \cos ^2\left(\frac{x+y}{2}\right)-4 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\)\(+1=0\)
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 \(\theta]\)
\(\begin{aligned}
& \Rightarrow \frac{x-y}{2}=0 \\
& \Rightarrow x=y
\end{aligned}\)