solve for $x$: $\frac{\sin(x)}{x}=\frac{5}{6}$ Is it possible to solve for x the following equation without root finding:
$$\frac{\sin(x)}{x}=\frac{5}{6}$$
 A: 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
$$ x\approx \sqrt{\frac{2}{977} \left(13405-\sqrt{166152805}\right)}.$$
A: $$x=6/5\cdot\sin(6/5\cdot\sin(6/5\cdot \sin(6/5\cdot\sin(\cdots))))$$
But from convergence point of view Newton is faster
A: 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 $$x_{(2)}=\frac{1}{3} \left(-\frac{2}{\sqrt{3}}+\pi +\sqrt{\frac{2}{3} \left(29-5 \sqrt{3}
   \pi \right)}\right)\approx 1.02674$$
