Solving in $x$ the equation$ \frac{l}{d}=\frac{x}{\sin(x)}$

Question

Is there a way to solve for $x$ in this equation? $$\frac{l}{d}=\frac{x}{\sin(x)}$$ I was able to approximate it with a Taylor series expansion of $\sin(x)$, but its really bugging me that I can't solve for x exactly.

If you can't solve for $x$ (without any kind of approximation), why?

Answer 1

This is simply $x=\operatorname{sinc}^{-1}(d/l)$, which doesn't have a more elementary form aside from trivial solutions. Shortly put, the two $x$'s here are on two different 'levels', and it's impossible to put them on the same level since the only ways you can manipulate them with trig functions/identities will always leave at least one $x$ in either a trig or inverse trig function.

Answer 2

Let $k=\frac l d$ and consider the equation $$k=\frac{x}{\sin(x)}$$ You cannot solve it explicitly but you can have a rather good approximation of the solution before starting a numerical method.

For example, using the approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ (proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician) you would face a quadratic equation $$\pi\left(\frac{5 \pi }{4}-4 k\right)+(4 k-\pi ) x+x^2=0\implies x=\sqrt{4 k^2+2 \pi k-\pi ^2}-2 k+\frac{\pi }{2}$$

Starting with estimate as $x_0$, Newton method would converge in a couple of iterations $$x_{n+1}=k\,\frac{ x_n \cos (x_n)- \sin (x_n)}{k \cos (x_n)-1}$$

Let us try $$\left( \begin{array}{cccc} k & x_0 & x_1 & \text{solution} \\ 1.1 & 0.74262 & 0.74907 & 0.74899 \\ 1.2 & 1.02288 & 1.02676 & 1.02674 \\ 1.3 & 1.21992 & 1.22150 & 1.22150 \\ 1.4 & 1.37211 & 1.37259 & 1.37259 \\ 1.5 & 1.49572 & 1.49578 & 1.49578 \\ 1.6 & 1.59934 & 1.59935 & 1.59935 \\ 1.7 & 1.68815 & 1.68828 & 1.68828 \\ 1.8 & 1.76555 & 1.76586 & 1.76586 \\ 1.9 & 1.83386 & 1.83438 & 1.83438 \\ 2.0 & 1.89477 & 1.89549 & 1.89549 \\ 2.1 & 1.94956 & 1.95046 & 1.95046 \\ 2.2 & 1.99919 & 2.00024 & 2.00024 \\ 2.3 & 2.04443 & 2.04559 & 2.04559 \\ 2.4 & 2.08587 & 2.08713 & 2.08713 \\ 2.5 & 2.12403 & 2.12535 & 2.12535 \\ 2.6 & 2.15929 & 2.16065 & 2.16065 \\ 2.7 & 2.19201 & 2.19339 & 2.19339 \\ 2.8 & 2.22246 & 2.22385 & 2.22385 \\ 2.9 & 2.25089 & 2.25227 & 2.25227 \\ 3.0 & 2.27751 & 2.27886 & 2.27886 \end{array} \right)$$

Another simple approximation could be obtained after building the $[4,4]$ Padé approximant of the rhs. This would give $$k=\frac{\frac{5 }{11088}x^4+\frac{13 }{396}x^2+1}{\frac{551 }{166320}x^4-\frac{53 }{396}x^2+1}$$ which is a quadratic equation in $x^2$.