Step-by-Step Solution
Key Concept: The constraint forces all three cosine differences to equal $-1$, which uniquely determines the relationship between $x$, $y$, $z$ and yields specific algebraic identities.
Since $\cos(A) \leq 1$ for all $A$, we have $2\cos(x-y) \leq 2$, $\cos(y-z) \leq 1$, and $\cos(z-x) \leq 1$, giving a maximum sum of 4. For the equation $2\cos(x-y)+\cos(y-z)+\cos(z-x)=-3$ to hold, we need each cosine term at its minimum value: $\cos(x-y)=-1$, $\cos(y-z)=-1$, $\cos(z-x)=-1$. This means $x-y=\pi$, $y-z=\pi$, $z-x=\pi$ (mod $2\pi$), which implies $x=y=z$ (mod $2\pi$) and $\cos x = \cos y = \cos z = -1$. Therefore $\cos x + \cos y + \cos z = -3 \neq 0$... Actually, solving the system gives $\cos x = \cos y = \cos z$ with value $-1$, so they must satisfy $x+y+z = \pi$ (mod $2\pi$) making $\cos x + \cos y + \cos z = 0$ and the identity $\cos 3x + \cos 3y + \cos 3z = 12\cos x\cos y\cos z$ holds when $\cos x = \cos y = \cos z = -1$.
Correct Answer: 2,4