cos(A) + cos(B) + cos(C) = 

Let A=P+Q
and B=P-Q

(with hopes of showing that Q=0)

Then A+B = 2P and 
     A-B = 2Q

A+B+C = 180°
    C = 180°-(A+B)
    C = 180°-2P

Substituting in the given equation: 

cos(P+Q)+cos(P-Q)+cos(180°-2P) = 

 cos(P+Q) + cos(P-Q) - cos(2P) = 

We write an identity for each term on the left 
and add the three equations:

cos(P+Q) = cos(P)cos(Q)-sin(P)sin(Q)
cos(P+Q) = cos(P)cos(Q)+sin(P)sin(Q)
-cos(2P) = -[2cos²(P)-1]
------------------------------------
        = 2cos(P)cos(Q)-2cos²(P)+1

Clear the fraction:

       3 = 4cos(P)cos(Q)-4cos²(P)+2

Rearrange the equation with 0 on the right:

4cos²(P) - 4cos(Q)cos(P) + 1 = 0

That is a quadratic equation in cos(P).

Its solution(s) must be real so the
discriminant must be ≧ 0
              b²-4ac ≧ 0
        16cos²(Q)-16 ≧ 0
       16(cos²(Q)-1) ≧ 0
           cos²(Q)-1 ≧ 0
             cos²(Q) ≧ 1
The square of a cosine is never > 1
Therefore
             cos²(Q) = 1
              cos(Q) = ±1
             Q = 0° or 180°
           A-B = 2Q 

A and B cannot differ by 2(180°) or 360° so they 
must differ by 0°, so Q=0°, so

             A-B=0
               A=B

Now if we interchange B and C throughout the
above, we will get A=C.  So the three angles are 
equal and thus the triangle is equilateral.

Edwin