Angles satisfy this sole equation!

First, we do some manipulation for the $LHS$ $LHS=1+2[cos(A+B)+cos(A-B)]cosC = 1-2cos^2C-2cos(A+B)cos(A-B)=-2cos2A - 2cos2B-2cos2C$ For the $RHS$ , we prove this following inequality $cosA+cosB+cosC\leq\frac{3}{2}$ We have $cosA+cosB+cosC=-2cos^2\big(\frac{A+B}{2}\big)+2cos\big(\frac{A-B}{2}\big)cos\big(\frac{A+B}{2}\big)+1$ , so by setting $t=cos\big(\frac{A+B}{2}\big)$ , we'll get $f(t)=-2t^2+2cos\bigg(\frac{A-B}{2}\bigg)t+1\leq\frac{cos^2\big(\frac{A-B}{2}\big)+2}{2}\leq\frac{3}{2}$ $\therefore RHS=LHS\leq\frac{3}{2}$ $\Leftrightarrow -2(cos2A+cos2B+cos2C)\leq\frac{3}{2}$ Through some manipulation, we'll get the above inequality as $cos^2A+cos^2B+cos^2C\geq\frac{3}{4} \ (2)$ . So for the finishing touch, we need to prove $cos^2A+cos^2B+cos^2C\leq\frac{3}{4}$ Without losing generality, we assume that $A\geq B\geq C$ , then we'll have $C\leq 60^{\circ}$ and $A+B\geq 120^{\circ}$ . Now for the LHS $cos^2A+cos^2B+cos^2C=cos^2(A+B)+cos(A-B)cos(A+B)+1$ Since $A+B\geq 120^{\circ}$ , $cos(A+B)\leq\frac{-1}{2}$ , applying this to the $LHS$ $cos^2(A+B)+cos(A-B)cos(A+B)+1\leq cos^2(A+B)+cos(A+B)+1\leq \frac{1}{4}-\frac{1}{2}+1=\frac{3}{4}$ So (2) only happens when $\begin{cases} A+B=120^{\circ} \\ A-B=0^{\circ} \\ A+B+C=180^{\circ}\end{cases}$ , therefore $A=B=C$ and we obtain the equilateral triangle

$LHS=1+4cosAcosBcosC$ $=1+2(cos(A-B)+cos(A+B))cosC$ $=1+2cos(A-B)cosC+2cos(A+B)cosC$ $=1+cos(A-B-C)+cos(A-B+C)+cos(A+B-C)+cos(A+B+C)$ But, $cos(A+B+C)=cos\pi=-1$ So, $LHS=cos(A+B-C)+cos(B+C-A)+cos(C+A-B)$ Now, this left hand side can be equal to the right hand side (which is $cosA+cosB+cosC$ ) if we set, $A=B=C$ . Hence, the triangle is equilateral.

P.S. There's another way it seems! $LHS=\sum\limits_{A,B,C}^{cyc} cos(A+B-C)=\sum\limits_{A,B,C}^{cyc} cos(A+B+C-2C)$ $=\sum\limits_{A,B,C} cos(\pi-2C)=\sum\limits_{A,B,C} (-cos(2C))$ $=\sum\limits_{A,B,C} (1-2cos^2 C)$ We can then set each term on the LHS equal to the corresponding term on the RHS. $1-2cos^2 C=cos C$ Solving the above equation gives us $C=\frac{\pi}{3}$ . Similarly, $A=B=\frac{\pi}{3}$ .