$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ 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.

$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
$$\frac D4= \cos^2 b \cdot \cos^2c-\cos^2b -\cos^2c+1= \left( 1-\cos^2b\right)\left(1-\cos^2 c \right)=\sin^2b \cdot \sin^2c$$
$$x_{1,2}=-\cos b \cdot \cos c \pm \sin b \cdot \sin c=-\cos (b \mp c)$$
$$x_1=-\cos (b+c)=-\cos (\pi - a)=\cos a$$
$$x_2=-\cos(b-c)$$
