<s>[INST] $x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ 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.
 [/INST] "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.
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