If $\cos A+\cos B+\cos C=\frac{3}{2}$,  then show that the triangle is equilateral.

Given $\cos A+\cos B+\cos C=\frac{3}{2}$

$2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\cos C=\frac{3}{2}$

$2\cos \left(\frac{\pi }{2}-\frac{c}{2}\right)\cos \left(\frac{A-B}{2}\right)+\cos C=\frac{3}{2}$

$2\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}=\frac{3}{2}$

$1+2\sin \frac{C}{2}\left(\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right)=\frac{3}{2}$

$1+2\sin \frac{C}{2}\left(\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right)=\frac{3}{2}$

$1+2\sin \frac{C}{2}(2\sin A/2\sin B/2)=\frac{3}{2}$

$1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{2}$

$\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{8}=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$

$\begin{array}{r}\sin \frac{A}{2}=\frac{1}{2}\\ \frac{A}{2}={30}^{\circ }\\ \mathrm{A}={60}^{\circ }\end{array}\left|\begin{array}{l}\sin \frac{B}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{1}{2}\\ \frac{B}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}={30}^{\circ }\\ \text{ B }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}={60}^{\circ }\end{array}\right|\begin{array}{r}\sin \frac{C}{2}=\frac{1}{2}\\ \frac{C}{2}={30}^0\\ C={60}^0\end{array}$

$\Rightarrow A=B=C$

$\therefore \mathrm{△}ABC$ is an Equilateral triangle