Paso 1: Tomar $x=45^\circ$:
$$ s=u=\frac{\sqrt2}{2},\qquad \frac{su}{s+u-1}=\frac{\frac12}{\sqrt2-1}=\frac{1+\sqrt2}{2}. $$
Entonces:
$$ a\frac{\sqrt2}{2}+b\frac{\sqrt2}{2}+c=\frac{1+\sqrt2}{2}. $$

Paso 2: Tomar $x=60^\circ$:
$$ s=\frac{\sqrt3}{2},\;u=\frac12,\; \frac{su}{s+u-1}=\frac{\frac{\sqrt3}{4}}{\frac{\sqrt3}{2}+\frac12-1} =\frac{\frac{\sqrt3}{4}}{\frac{\sqrt3-1}{2}}=\frac{\sqrt3}{2(\sqrt3-1)}=\frac{3+\sqrt3}{4}. $$
Luego:
$$ a\frac{\sqrt3}{2}+b\frac12+c=\frac{3+\sqrt3}{4}. $$

Paso 3: Tomar $x=120^\circ $:
$$ s=\frac{\sqrt3}{2},\;u=-\frac12,\; \frac{su}{s+u-1}=\frac{-\frac{\sqrt3}{4}}{\frac{\sqrt3}{2}-\frac12-1} =\frac{-\frac{\sqrt3}{4}}{\frac{\sqrt3-3}{2}}=\frac{\sqrt3}{2(3-\sqrt3)}=\frac{1+\sqrt3}{4}. $$
Así:
$$ a\frac{\sqrt3}{2}-b\frac12+c=\frac{1+\sqrt3}{4}. $$

Paso 4: Restar las dos últimas ecuaciones para hallar $ b $:
$$ \left(a\frac{\sqrt3}{2}+b\frac12+c\right)-\left(a\frac{\sqrt3}{2}-b\frac12+c\right)=\frac{3+\sqrt3}{4}-\frac{1+\sqrt3}{4} \Rightarrow b= \frac12. $$
Sumándolas:
$$ a\sqrt3+2c=\frac{(3+\sqrt3)+(1+\sqrt3)}{4}=\frac{4+2\sqrt3}{4}=1+\frac{\sqrt3}{2} \Rightarrow a\sqrt3+2c=1+\frac{\sqrt3}{2}. $$
De la ecuación con $45^\circ$ y $b=\frac12$:
$$ \frac{\sqrt2}{2}a+\frac{\sqrt2}{2}\cdot\frac12+c=\frac{1+\sqrt2}{2} \Rightarrow \frac{\sqrt2}{2}a+c=\frac12. $$
Resolviendo, se obtiene:
$$ a=\frac12,\qquad b=\frac12,\qquad c=\frac12. $$

Resultado final:
$$ a+b+c=\frac12+\frac12+\frac12=\boxed{\frac32}. $$