Is there any way to rearrange the following equation for x?

Question

Is there any way to rearrange the following equation to get $x$ in terms of $\theta$? $g$, $k$ and $u$ are known constants, and $\theta$ and $x$ are unknowns.

$$\frac g{k^2}\ln(\frac {u \cos(\theta)-kx}{u \cos(\theta)})+x\tan(\theta)+\frac{gx}{ku\cos(\theta)}=0$$

My ultimate aim is to differentiate $x$ with respect to $\theta$, equate that expression to $0$, and thereby find a value for $\theta$ that maximises $x$; if there is a way to do this without rearranging for $x$, I would be very interested to hear it.

I have briefly looked into the Lambert W function, but as far as I can tell it can't be used here, as $\theta$ is an unknown.

Edit: I took $\frac {d}{d\theta}$, and got the following equation (rearranged for $x$). Can I use this to find the optimal value of $\theta$?

$$x=\frac {u^2\cos(\theta)}{ku+g\sin(\theta)}$$

Answer 1

Indeed,

$$\frac{df(\theta,x)}{d\theta}=\frac{\partial f(\theta,x)}{\partial\theta}+\frac{dx}{d\theta}\frac{\partial f(\theta,x)}{\partial x}=0$$ so that $\dfrac{dx}{d\theta}$ cancels when $\dfrac{\partial f(\theta,x)}{\partial\theta}$ cancels.

I confirm the result

$$x=\frac {u^2\cos(\theta)}{ku+g\sin(\theta)},$$

but you still have to solve the initial equation for $\theta$, after plugging the expression of $x$. Which is ugly.

Answer 2

In case anyone stumbles upon this in the future:

I have spoken to my teacher, who recommended looking into implicit differentiation. Using this, we can solve for x, and plug it back into the original equation and solve for θ