Certainly! Let's solve for [tex]\(\cos A\)[/tex], [tex]\(\cos B\)[/tex], and [tex]\(\cos C\)[/tex] given the system of equations:
[tex]\[
\begin{cases}
a = b \cos C + c \cos B \\
b = c \cos A + a \cos C \\
c = a \cos B + b \cos A \\
\end{cases}
\][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are non-zero.
To solve this system of equations, we can use the symmetry and properties of the cosine function. Observe that the equations are symmetric with respect to [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].
Let's denote the following:
[tex]\[
\begin{cases}
a = b \cos C + c \cos B \\
b = c \cos A + a \cos C \\
c = a \cos B + b \cos A \\
\end{cases}
\][/tex]
Let's assume without loss of generality that [tex]\(a = b = c = 1\)[/tex]. This simplifies our equations to:
[tex]\[
\begin{cases}
1 = 1 \cos C + 1 \cos B \\
1 = 1 \cos A + 1 \cos C \\
1 = 1 \cos B + 1 \cos A \\
\end{cases}
\][/tex]
This simplifies further to:
[tex]\[
\begin{cases}
1 = \cos C + \cos B \\
1 = \cos A + \cos C \\
1 = \cos B + \cos A \\
\end{cases}
\][/tex]
Adding all three equations, we have:
[tex]\[
1 + 1 + 1 = (\cos C + \cos B) + (\cos A + \cos C) + (\cos B + \cos A)
\][/tex]
This simplifies to:
[tex]\[
3 = 2(\cos A + \cos B + \cos C)
\][/tex]
Therefore, we can solve for [tex]\(\cos A + \cos B + \cos C\)[/tex]:
[tex]\[
3 = 2(\cos A + \cos B + \cos C)
\][/tex]
[tex]\[
\cos A + \cos B + \cos C = \frac{3}{2}
\][/tex]
Next, we substitute back into one of our individual equations. From the first equation, substituting [tex]\(\cos C + \cos B = 1\)[/tex], we get:
[tex]\[
\cos A = 1 - \cos C
\][/tex]
However, using the fact that the equations are symmetrical, we see that each of the cosine terms must be equal, i.e.:
[tex]\[
\cos A = \cos B = \cos C
\][/tex]
Substituting [tex]\(\cos A = \cos B = \cos C\)[/tex] into the sum equation:
[tex]\[
3 \cos A = \frac{3}{2}
\][/tex]
Solving for [tex]\(\cos A\)[/tex], we get:
[tex]\[
\cos A = \frac{1}{2}
\][/tex]
Therefore, we find:
[tex]\[
\cos A = \cos B = \cos C = \frac{1}{2}
\][/tex]
Thus, the solutions are:
[tex]\[
\boxed{\cos A = \cos B = \cos C = \frac{1}{2}}
\][/tex]
