Question 883143
<pre>
{{{cos(x)+cos(y)-cos(x+y)=3/2}}}

{{{cos(x)+cos(y)-(cos(x)cos(y)-sin(x)sin(y))=3/2}}}

{{{cos(x)+cos(y)-cos(x)cos(y)+sin(x)sin(y)=3/2}}}

{{{cos(x)+cos(y)-cos(x)cos(y) +- sqrt(1-cos^2(x))sqrt(1-cos^2(y))=3/2}}}

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

{{{a+b-ab +- sqrt(1-a^2)sqrt(1-b^2)=3/2}}}

{{{a+b-ab -3/2 = "" +- sqrt(1-a^2)sqrt(1-b^2)}}

{{{(a+b-ab -3/2)^2 = (1-a^2)(1-b^2)}}}

{{{(a+b-ab -3/2)^2 = (1-a^2)(1-b^2)}}}

{{{a^2+b^2+a^2b^2+9/4+2ab-2a^2b-3a+2ab^2-3b+3ab = 1-b^2-a^2+b^2}}}

{{{2a^2b-2a^2+2ab^2-5ab+3a-2b^2+3b-5/4 = 0}}}

{{{8a^2b-8a^2+8ab^2-20ab+12a-8b^2+12b-5 = 0}}}

{{{8a^2b-8a^2+8ab^2-20ab+12a-8b^2+12b-5 = 0}}}

{{{(8b-8)a^2+(8b^2-20b+12)a+(-8b^2+12b-5) = 0}}}

{{{8(b-1)a^2+4(2b^2-5b+3)a+(-8b^2+12b-5) = 0}}}

{{{8(b-1)a^2+4(2b-3)(b-1)a+(-8b^2+12b-5) = 0}}}

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

{{{4^2(2b-3)^2(b-1)^2-4*8(b-1)(-8b^2+12b-5)>=0}}}

{{{16(2b-3)^2(b-1)^2-32(b-1)(-8b^2+12b-5)>=0}}}

{{{(2b-3)^2(b-1)^2-2(b-1)(-8b^2+12b-5)>=0}}}

{{{(b-1)((2b-3)^2(b-1)-2(-8b^2+12b-5))>=0}}}

{{{(b-1)((4b^2-12b+9)(b-1)+16b^2-24b+10)>=0}}}

{{{(b-1)(4b^3-12b^2+9b-4b^2+12b-9+16b^2-24b+10)>=0}}}

{{{(b-1)(4b^3-3b+1)>=0}}}

We factor the cubic in the parentheses. 
It has possible rational zeros {{{"" +- 1}}}, {{{"" +- 1/2}}}, {{{"" +- 1/4}}}.

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

-1|4  0 -3  1
  |<u>  -4  4 -1</u>
   4 -4  1  0

So we have further factored the expression 
on the left as

{{{(b-1)(b+1)(4b^2-4b+1)>=0}}} 

and we factor once more as

{{{(b-1)(b+1)(2b-1)^2>=0}}}

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:

<============&#9679;-----&#9679;-&#9679;===========>
-4  -3  -2  -1   0 {{{1/2}}}1   2   3   4

({{{-infinity}}},-1] U {{{"{"}}}{{{1/2}}} }  U [1,{{{infinity}}})

Since b = cos(x), and {{{-1<=cos(x)<=1}}}, we have 

cos(x)=-1, cos(x)={{{1/2}}}, cos(x)=1
     x={{{pi}}},       x={{{pi/3}}},     x=0

The only possibility for x is {{{pi/3}}}.

Interchange x and y in the above and 
we get y = {{{pi/3}}}      

Interchange x and z in the above and 
we get z = {{{pi/3}}}.

So all three angles are {{{pi/3}}} and so
the triangle is equilateral. 

Edwin</pre>