System differential equations 
Solve the system differential equation
$$\frac{dx}{\cos y}=\frac{dy}{\cos x}=\frac{dz}{\cos x \cos y}$$

I think:
$$\frac{dx}{\cos y}=\frac{dy}{\cos x}$$
$$\cos x~dx = \cos y~dy$$
$$\sin x = \sin y + C_1$$
$$C_1 = \sin x - \sin y$$
And then I do not know how to solve it.

Maybe:
$$\frac{dxdy}{\cos y \cos x}=\frac{dz}{\cos x \cos y}$$
$$dx dy=dz$$
But what next?
 A: Probably, the system of ODEs comes from solving of a PDE such as :
$$\cos(y) z_x+\cos(x) z_y =\cos(x) \cos(y) $$
where the unknown function is $z(x,y)$.
$$\frac{dx}{\cos y}=\frac{dy}{\cos x}=\frac{dz}{\cos x \cos y}$$
The equation of a first family of characteristic curves comes from :
$$\frac{dx}{\cos y}=\frac{dy}{\cos x} \quad\to\quad \sin(x)-\sin(y)=C_1$$
That is what you did.
The equation of a second family of characteristic curves comes from :
$$\frac{dy}{\cos x}=\frac{dz}{\cos x \cos y} \quad\to\quad dz=\cos(y)dy \quad\to\quad z+\sin(y)=C_2$$
The general solution of the PDE can be expressed on various equivalent forms, for example :
$$\Phi\left(z+\sin(y)\:,\:\sin(x)-\sin(y)\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Or explicitly :
$$z+\sin(y)=F\left(\sin(x)-\sin(y)\right)\quad\to\quad z=-\sin(y)+F\left(\sin(x)-\sin(y)\right)$$
where $F$ is any differentiable function.
Other equivalent forms are possible.
NOTE :
Alternatively, one can chose the other family of characteristics curves, from :
$$\frac{dx}{\cos y}=\frac{dz}{\cos x \cos y} \quad\to\quad \cos(x)dx=dz \quad\to\quad z+\sin(x)=C_3$$
This leads to 
$$z+\sin(x)=G\left(\sin(x)-\sin(y)\right)\quad\to\quad z=-\sin(x)+G\left(\sin(x)-\sin(y)\right)$$
where $G$ is any differentiable function.
This is the same general solution than above since the functions $F$ and $G$ are related. $\quad G\left(\sin(x)-\sin(y)\right)=\left(\sin(x)-\sin(y)\right)+F\left(\sin(x)-\sin(y)\right)$.
A: The first step is ok, it says
$$\frac{dx}{\cos(y)}=\frac{dy}{\cos(x)}$$
and you have
$$\sin(x)=\sin(y)+c_{1}$$
Then you have also
$$\frac{dy}{\cos(x)}=\frac{dz}{\cos(x)\cos(y)}$$
This gives
$$dz=\cos(y)dy$$
or
$$z=\sin(y)+c_{2}$$
Equivalently
$$\frac{dx}{\cos(y)}=\frac{dz}{\cos(x)\cos(y)}$$
or
$$dz=\cos(x)dx$$
This gives
$$z=\sin(x)+c_{3}$$
And using the first result
$$\sin(x)=\sin(y)+c_{1}$$
you have $$c_{1}+c_{2}=c_{3}$$
