Can anyone please help me to find the generalized equation for $x$,$y$ and $z$?

Question

$$ A= P(9+2\cos(x)+2\cos(y)+2\cos(z)) $$ $$ B= P(9+2\cos(x-2\pi/3)+2\cos(y-2\pi/3)+2\cos(z-2\pi/3)) $$ $$ C= P(9+2\cos(x+2\pi/3)+2\cos(y+2\pi/3)+2\cos(z+2\pi/3)) $$

Answer 1

Observation: replacing $A,B,C$ with $A/p$, etc., we have \begin{align} A&= (9+2\cos(x)+2\cos(y)+2\cos(z))\\ B&= (9+2\cos(x-2\pi/3)+2\cos(y-2\pi/3)+2\cos(z-2\pi/3))\\ C&= (9+2\cos(x+2\pi/3)+2\cos(y+2\pi/3)+2\cos(z+2\pi/3)) \end{align}

Summing, we get \begin{align} A+B+C&= 27 + 2[\cos(x) + \cos(x-2\pi/3) + \cos(x + 2\pi/3] + \ldots \end{align} where the ellipses denote similar expressions in $y$ and $z$. But this expression in brackets is always zero, so these equations cannot have a solution unless $A + B + C = 27$.

Answer 2

Hint: Like John Hughes said first introduce $A'=A/p$, $B'=B/p$ and $C'=C/p$ assuming $p\neq0$. Only if $A=B=C=0$ allows $p=0$ which means that any combination of $x,y,z$ solves the problem.

Then use the addition formula

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

Then replace $\sin(x)=\sqrt{1-\cos^2(x)}$ (assuming $x \in [0,\pi]$), same for $y$ and $z$. And substitute $\cos(x)=\alpha_1$. Do the same with $\cos(y)=\alpha_2$ and $\cos(z)=\alpha_3$. Solve the nonlinear system.