Let a = cos(x), b = cos(y) 





















That's a quadratic in a, and the solutions must be
real, so the discriminant must be non-negative, so















We factor the cubic in the parentheses. 
It has possible rational zeros , , .

It's easy to see that -1 is a zero.  So

-1|4  0 -3  1
  |  -4  4 -1
   4 -4  1  0

So we have further factored the expression 
on the left as

 

and we factor once more as



The critical numbers are the zeros of the
left side, which are 1, -1, and 1/2

Put those on a number line, test the intervals,
and the critical values, and get the graph of 
solution inequality as:

<============●-----●-●===========>
-4  -3  -2  -1   0 1   2   3   4

(,-1] U  }  U [1,)

Since b = cos(x), and , we have 

cos(x)=-1, cos(x)=, cos(x)=1
     x=,       x=,     x=0

The only possibility for x is .

Interchange x and y in the above and 
we get y =       

Interchange x and z in the above and 
we get z = .

So all three angles are  and so
the triangle is equilateral. 

Edwin