solve for $x$: $\frac{\sin(x)}{x}=\frac{5}{6}$ Is it possible to solve for x the following equation without root finding:
$$\frac{\sin(x)}{x}=\frac{5}{6}$$

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 $$x_{(2)}=\frac{1}{3} \left(-\frac{2}{\sqrt{3}}+\pi +\sqrt{\frac{2}{3} \left(29-5 \sqrt{3}
   \pi \right)}\right)\approx 1.02674$$
