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Is it possible to solve for x the following equation without root finding:

$$\frac{\sin(x)}{x}=\frac{5}{6}$$

Christopher Carl Heckman
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ZeroCool
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    Wolfram alpha just does it numerically...getting $x\sim \pm 1.02674$. I expect there's no simpler form. – lulu May 14 '16 at 23:27
  • I'm surprised how challenging such simple equations! – ZeroCool May 14 '16 at 23:35
  • Exactly? Don't think so. But there are plenty of nice numerical methods, Taylor series approximations and so on. But you're aware of that already. – KR136 May 14 '16 at 23:39
  • It's not as surprising as you might think. There's a lot of numbers out there. Not all of them are as simple as, let's say, solutions to polynomials with integer coefficients. – Lee Mosher May 14 '16 at 23:41
  • If you assume small angles then you can obtain $x=\pm 1$ fairly fast else it's a bit grim as stated. You can't avoid finding roots though because even to lowest order the LHS expansion is $1 - x^2/3! + \mathcal{O}(x^4)$ – Alexander McFarlane May 15 '16 at 00:40
  • Does anyone know how to show that this equation doesn't have a closed form solution using standard functions, or some such similar result? (The answers so far seem to do exactly what the OP asks them not to do: use some form of root finding.) – Simon S May 15 '16 at 02:20
  • @SimonS: that depends on the meaning of closed form. By Lagrange's inversion formula, we may solve $\frac{\sin x}{x}=\frac{5}{6}$ in terms of a pretty complicated series. Is that a closed form? – Jack D'Aurizio May 15 '16 at 03:59
  • By closed form I mean it is possible to write an expression for $x$ that requires a finite number of calculations using combinations of standard functions; for instance, the expression for $x_0$ in your answer below. However, the OP may have a different definition. – Simon S May 15 '16 at 05:03

3 Answers3

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There is no closed form solution, but since $\sin(x)$ is a concave function on $(0,\pi)$ and $\sin'(0)=1$, there is a unique real and positive solution to $\sin(x)=\frac{5}{6}x$. We may use a step of Newton's method with starting point $x_0=1$ get a tight approximation of such solution: $$ x_1\approx 1-\frac{\sin 1-5/6}{\cos 1-5/6}\approx 1.02777. $$ An even better starting point is given by the solution of $$ \exp\left(\frac{5x^2}{x^2-30}\right)=\frac{5}{6} $$ i.e. by $$ x_0 = \sqrt{\frac{30 \log\left(\frac{6}{5}\right)}{5+\log\left(\frac{6}{5}\right)}}\approx 1.0273477$$ or by the solution of $$ \frac{60-7x^2}{60+3x^2}=\frac{5}{6}, $$ i.e. by $$ x_0 = 2\sqrt{\frac{5}{19}} \approx 1.026. $$ By using the Padé approximant in the origin $\frac{p(x)}{q(x)}$ with $\partial p=\partial q=4$, we get the almost exact solution $$ x\approx \sqrt{\frac{2}{977} \left(13405-\sqrt{166152805}\right)}.$$

Jack D'Aurizio
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$$x=6/5\cdot\sin(6/5\cdot\sin(6/5\cdot \sin(6/5\cdot\sin(\cdots))))$$

But from convergence point of view Newton is faster

MrYouMath
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Equations which mix polynomial and trigonometric terms do not show analytical solutions (this is already the case for $x=\cos(x)$) and numerical methods should be used.

However, some rather good approximations can be made and, for your curiosity, I give you the links to two questions of mine (here and here).

Using the beautiful $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician, your equation becomes $$\frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}=\frac 56 x$$ which reduces to a simple quadratic $$-\frac{10 x^2}{3}+\left(\frac{10 \pi }{3}-16\right) x-\frac{25 \pi ^2}{6}+16 \pi=0$$ for which the solution to keep is $$x=\frac{1}{40} \left(-96+20 \pi +\sqrt{9216+3840 \pi -1600 \pi ^2}\right)\approx 1.02288 $$ while the exact solution, obtained using Newton method, would be $\approx 1.02674$.

Now, knowing that the solution is close to $1$, we could approximate $\sin(x)$ using its simplest Padé approximant built at $x=1$; this should give $$\sin(x)=\frac{\frac{1}{2} (x-1) (2 \cos (1)+\sin (1) \tan (1))+\sin (1)}{1+\frac{1}{2} (x-1) \tan (1)}$$ which, again, reduces to a quadratic the solution of which being $\approx 1.02676$.

Another way could be to use the Taylor series of $\sin(x)$ around $x=\frac \pi 3$ which is $$\sin(x)=\frac{\sqrt{3}}{2}+\frac{1}{2} \left(x-\frac{\pi }{3}\right)-\frac{1}{4} \sqrt{3} \left(x-\frac{\pi }{3}\right)^2-\frac{1}{12} \left(x-\frac{\pi }{3}\right)^3+O\left(\left(x-\frac{\pi }{3}\right)^4\right)$$ Limiting to first order, the solution would then be $$x_{(1)}=\frac{1}{2} \left(3 \sqrt{3}-\pi \right)\approx 1.02728$$ Limiting to second order, the solution would then be $$x_{(2)}=\frac{1}{3} \left(-\frac{2}{\sqrt{3}}+\pi +\sqrt{\frac{2}{3} \left(29-5 \sqrt{3} \pi \right)}\right)\approx 1.02674$$

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Claude Leibovici
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