$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ 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.
 A: "Completing the square" would be my choice.
$x^2+ (2cos(b)cos(c))x+ cos^2(b)cos^2(c)- cos^2(b)cos^2(c)+ cos^2(b)+ cos^2(c)- 1= 0$
$(x+ cos(b)cos(c))^2= 1- cos^2(b)- cos^2(c)+ cos^2(b)cos^2(c)$
You say that "$a+ b+ c= \pi$ but there is NO "a" in your equation.
A: $$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
$$\frac D4= \cos^2 b \cdot \cos^2c-\cos^2b -\cos^2c+1= \left( 1-\cos^2b\right)\left(1-\cos^2 c \right)=\sin^2b \cdot \sin^2c$$
$$x_{1,2}=-\cos b \cdot \cos c \pm \sin b \cdot \sin c=-\cos (b \mp c)$$
$$x_1=-\cos (b+c)=-\cos (\pi - a)=\cos a$$
$$x_2=-\cos(b-c)$$
A: 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 $\cos^2\theta + \sin^2\theta = 1$:
$$ x = \frac{-2\cos b\cos c\pm 2\sqrt{(\cos^2 b -1)(\cos^2 c -1)}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm 2\sqrt{\sin^2b\sin^2c}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm 2\sin b\sin c}{2}  = -\cos b\cos c\pm \sin b\sin c $$
$$ x = (-\cos b\cos c+ \sin b\sin c) \ \text{or} \ (-\cos b\cos c- \sin b\sin c) $$
$$ x = -\cos(b+c)= \cos a \ \text{or} -\cos(b-c)$$
A: Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2B+\cos^2C-1=\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
$$x^2+2x\cos B\cos C+\cos^2B+\cos^2C-1=0$$
$$\iff x^2+x\{\cos(B+C)+\cos(B-C)\}+\cos(B+C)\cos(B-C)=0$$
Now use $y^2+(a+b)y+ab=(y+a)(y+b)$ and $\cos(B+C)=\cos(\pi-A)=-\cos A$
