# Is there a method for solving equations like $\sin x=x\cos(x)$ in closed form?

Is there a method for solving equations like $$\sin x=x\cos(x)$$ in closed form?

I was looking into involute curves and ran into two equations that I'd like to find closed form solutions to: $$\sin(t)=t\cos(t)$$ as well as $$\cos(t)=-t\sin(t)$$. I've messed around with these equations a bit numerically and found that the product of the first positive root of both equation is equal to about $$4\pi$$ if it matters. I tried to rearrange the equations and to apply trigonometric laws but I wasn't successful.

• Not sure if this helps but: since $cos$ and $sin$ cannot both be simultaneously zero, your equation is equivalent to solving $t = \tan(t)$ and $t = \cot(t)$. Commented Jul 12, 2021 at 21:33
• @AryamanMaithani: Over the reals, perhaps. Over the complex numbers, I tend to imagine that $\sin z - \cos z$ should have a zero somewhere (but I do not have a proof of that). Commented Jul 12, 2021 at 21:37
• Commented Jul 12, 2021 at 21:38
• The first positive solution of each equation are $4.4934...$ and $.8603...$ respectively. Its product $3.8658...$ it's nowhere near $4\pi$. You may want to check your calculations. Commented Jul 12, 2021 at 23:03
• @Kevin: $\sin - \cos$ has a zero even in the reals, $\pi/4$ for example. What I meant is that there's no $t$ for which $\sin t = \cos t = 0$. This remains true over $\Bbb C$ as well since $\sin^2 + \cos^2 = 1$. Commented Jul 13, 2021 at 6:19

As far as I know, you can only solve this numerically. It has infinite solutions, of course. It's equivalent to solving

$$x = \tan x,$$

and in physics it arises in a number of contexts that we want to solve

$$x = \tan k x.$$

for some real $$k$$. When it does, we usually just plot $$x, \tan k x$$ as functions of $$x$$ look to see where the curves intersect. You can use numerical methods to find different intersections. For some reasons $$x = \cot x$$ doesn't seem to arise as much, but it seems to function pretty similarly.

First check $$0$$ and $$1$$: $$0$$ is a solution of your first equation.

Neither of your two equations can be solved any further by elementary functions alone or by elementary functions and the special function Lambert W alone.

Let's take for example your equation from the title of your question:

$$\sin(x)=x\cos(x).$$

Rearrange the equation to have all functions of the solution variable on one side of the equation:

$$\sin(x)-x\cos(x)=0$$

Your equation is in dependence of $$x$$ and the transcendental functions $$\sin$$ and $$\cos$$ of $$x$$, and all coefficients of the equation are algebraic numbers. That means, the equation is an algebraic equation over the algebraic numbers in dependence of the solution variable and at least one transcendental function of the solution variable. Such types of equations often cannot be solved by elementary functions alone.

The left-hand side of the latter equation is an elementary function. The elementary functions can be generated by applying only $$\exp$$, $$\ln$$ and/or unary or multiary algebraic functions.
Each of the elementary standard functions (sin, cos, tan, cot, sec, csc, sinh, cosh, tanh, coth, sech, csch, arcsin, arccos, arctan, arccot, arcsec, arccsc, arcsinh, arccosh, arctanh, arccoth, arcsech, arccsch) can be brought to this expln-form. See e.g. the Wikipedia articles for the single functions or [Abramowitz/Stegun 1970]:

$$\sin(x)=-\frac{1}{2}i\left(e^{ix}-e^{-ix}\right),$$

$$\cos(x)=\frac{1}{2}\left(e^{ix}+e^{-ix}\right).$$

$$-\frac{1}{2}i\left(e^{ix}-e^{-ix}\right)-\frac{1}{2}x\left(e^{ix}+e^{-ix}\right)=0$$

$$-\frac{1}{2}x\left(e^{ix}\right)^2-\frac{1}{2}i\left(e^{ix}\right)^2-\frac{1}{2}x+\frac{1}{2}i=0$$

$$-x\left(e^{ix}\right)^2-i\left(e^{ix}\right)^2-x+i=0$$

$$x\rightarrow\frac{t}{i}:$$

$$it\left(e^{t}\right)^2-i\left(e^t\right)^2+it+i=0$$

$$t(e^t)^2-(e^t)^2+t+1=0\tag1$$

The equation is now in the expln-form.

[Abramowitz/Stegun 1970] Abramowitz, M.; Stegun, I.: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standard 1970

1.) Elementary inverses / elementary numbers

The elementary function on the left-hand side of this equation is an algebraic function over the algebraic numbers in dependence of $$t$$ and $$e^t$$. This algebraic function is therefore a multiary algebraic function over the algebraic numbers.
The theorem in [Ritt 1925], that is proved also in [Risch 1979], implies that bijective compositions of $$\exp$$, $$\ln$$ and/or unary algebraic functions are the only elementary functions that are invertible by an elementary function. The elementary function on the left-hand-side of equation 1 is not bijective because it is not injective, but you can split it into bijective pieces by choosing restrictions of this function (restricting the domain of the function). You are looking for the partial inverses then. But the elementary function on the left-hand side of equation 1 is not in the form that Ritt's theorem requires. If we cannot find a representation of the elementary function in this form, we therefore cannot solve the equation by only rearranging it by applying only elemenary functions that are readable from the equation.

2.) Elementary numbers

The elementary numbers are the numbers that are generated from the rational numbers by applying only elementary functions (or rather $$\exp$$, $$\ln$$ and/or unary or multiary algebraic functions).

Equation 1 is an irreducible polynomial equation in dependence of the solution variable $$t$$ and $$e^t$$ with only algebraic coefficients. The main theorem in [Lin 1983] says, assuming Schanuel's conjecture, that equations of this type cannot be solved by elementary numbers except $$0$$. [Chow 1999] proves, assuming Schanuel's conjecture, that equations of this type cannot be solved by explicit elementary numbers except $$0$$.

Although neither of your two equations is solvable by elementary numbers, products of solutions of these equations could perhaps be an elementary number like $$4\pi$$ is.

3.) Lambert W

For applying only Lambert W and elementary functions, equation 1 in dependence of $$t$$ and $$e^t$$ should be transformable to the form

$$f_1(a_1+a_2f_2(t)^{a_3}e^{b_1+b_2f_2(t)^{b_3}})=c\tag2,$$

where $$a_1,a_2,a_3,b_1,b_2,b_3,c$$ are constants and $$f_1$$ and $$f_2$$ are elementary functions with a suitable elementary partial inverse. Remember that Ritt's above mentioned theorem implies which types of elementary functions have elementary invertible partial inverses.
But your equation cannot be brought to this form.

4.) Generalized Lambert W

Your equations are solvable in terms of Generalized Lambert W.

$$\sin(t)=t\cos(t)$$ $$-\frac{1}{2}i(e^{it}-e^{-it})=\frac{1}{2}t(e^{it}+e^{-it})$$ $$t\to\frac{x}{2i}$$: $$e^x=-\frac{x+2}{x-2}$$ $$\frac{x-2}{x+2}e^x=-1$$ $$x=W\left(^{+2}_{-2};-1\right)$$ $$t=-\frac{1}{2}i\ W\left(^{+2}_{-2};-1\right)$$

$$\$$

$$\cos(t)=-t\sin(t)$$ $$\frac{1}{2}(e^{it}+e^{-it})=\frac{1}{2}it(e^{it}-e^{-it})$$ $$t\to\frac{x}{2i}$$: $$e^x=\frac{x+2}{x-2}$$ $$\frac{x-2}{x+2}e^x=1$$ $$x=W\left(^{+2}_{-2};1\right)$$ $$t=-\frac{1}{2}i\ W\left(^{+2}_{-2};+1\right)$$

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equa-tion. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generali-zed Lambert W functions. 2018
[Stoutemyer 2022] Stoutemyer, D. R.: Inverse spherical Bessel functions generalize Lambert W and solve similar equations containing trigonometric or hyperbolic subexpressions or their inverses. 2022