The first order linear PDE I am looking at it is.
$$(\cos(y) + \cos(x+y))z_x-(\cos(x)+\cos(x+y))z_y=0$$
When I try to solve for the characteristics, I obtain the following ODE's.
$$\frac{dx}{dt}=\cos(y)+\cos(x+y)$$ $$\frac{dy}{dt}=-\cos(x)-\cos(x+y)$$
I'm not really able to decouple the system, even through changing to diagonalized coordinates. Im not sure if I should expect the solutions for the characteristics to even have a closed form expression based on numerical results I get from these ODE's.
Dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ we obtain $$ \frac{dy}{dx}=-\frac{\cos(x)+\cos(x+y)}{\cos(y)+\cos(x+y)} $$ $$ \implies [\cos(x)+\cos(x+y)]dx+[\cos(y)+\cos(x+y)]dy=0. \tag{1} $$ Equation $(1)$ is an exact differential equation, since $$ \frac{\partial}{\partial y}[\cos(x)+\cos(x+y)]=-\sin(x+y)= \frac{\partial}{\partial x}[\cos(y)+\cos(x+y)]. \tag{2} $$ There is, therefore, a function $u(x,y)$ such that \begin{align} \frac{\partial u}{\partial x}&=\cos(x)+\cos(x+y),\tag{3a} \\ \frac{\partial u}{\partial y}&=\cos(y)+\cos(x+y). \tag{3b} \end{align} Integrating $(3\text{a})$ we find $$ u(x,y)=\sin(x)+\sin(x+y)+f(y). \tag{4} $$ From $(4)$ and $(3\text{b})$ it follows that $$ \frac{df}{dy}=\cos(y) \implies f(y)=\sin(y) + C. \tag{5} $$ Equation $(1)$ implies $du=0$, hence $u=$ constant, i.e. $$ \sin(x)+\sin(y)+\sin(x+y)=C_1. \tag{6} $$ The solution to the given PDE is, therefore, $$ z(x,y)=F(\sin(x)+\sin(y)+\sin(x+y)), \tag{7} $$ where $F$ is an arbitrary differentiable function.