You should simplify your equations - get rid of $\theta$. Next, use the symmetry of the situation to look at just one pulley moving towards the middle of the room. The other quantity we need is the slack in the rope - the difference between the width of the room and the length of the rope. Let's call that difference $2S$ (S for "slack").
The hypotenuse (rope hanging from pulley to weight) will have length $S+x$, and this results in a height $y$. So now we know
$$(S+x)^2 + x^2 = y^2\\
S^2 + 2Sx + 2x^2 = y^2\\
(2S+4x)dx = 2y dy\\
If we have the rope at a certain height $y$ for a given $x$, we can compute $S$. And now you have what you need so solve this (you don't need to solve for x and y; you just need the relationship - that is, the derivative of y with respect to x).