$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.

Not as elegant as Roman83's solution but here you go:
Solve the quadratic equation in $x$:
$$ x = \frac{-B\pm \sqrt{B^2 - 4AC}}{2A}$$
$A = 1, \ B = 2\cos b\cos c, \ C =\cos^2 b +\cos^2 c - 1 $
$$ x = \frac{-2\cos b\cos c\pm \sqrt{(2\cos b\cos c)^2 - 4(\cos^2 b +\cos^2 c - 1)}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm \sqrt{4(\cos^2 b\cos^2 c - \cos^2 b -\cos^2 c + 1)}}{2}$$
Using $\cos^2\theta + \sin^2\theta = 1$:
$$ x = \frac{-2\cos b\cos c\pm 2\sqrt{(\cos^2 b -1)(\cos^2 c -1)}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm 2\sqrt{\sin^2b\sin^2c}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm 2\sin b\sin c}{2}  = -\cos b\cos c\pm \sin b\sin c $$
$$ x = (-\cos b\cos c+ \sin b\sin c) \ \text{or} \ (-\cos b\cos c- \sin b\sin c) $$
$$ x = -\cos(b+c)= \cos a \ \text{or} -\cos(b-c)$$
