Number of solutions of $\sin x= \frac x {10}$ The easiest way to do this is plot graphs of $10 \sin x$ and $y=x$ to find number of intersection points.
However, is there a non graphical solution to this problem? 
Edit: 
My problem is different from the linked problem because the answers to that focus only on graphical solutions. 
 A: You cannot solve the equation exactly but you can very well find the number of solutions.
The extrema of the function $10\sin x-x$ are found at the roots of $10\cos x-1=0$, i.e. $\pm\arccos\dfrac1{10}+2k\pi$. The values at the extrema are thus
$$\pm10\sqrt{1-\frac1{100}}\mp\arccos\frac1{10}-2k\pi.$$
As the sine is bounded by $\pm1$, by solving $\pm10-x=0$ we know that no root can arise outside $[-10,10]$. By numerical computation of the extrema in that range (with a sufficient approximation such that the signs are guaranteed exact), you spot $7$ changes of signs, hence $7$ roots.
$$\begin{matrix}x& f(x)\\\hline
-10& \ge0\\
-7.8& -2.2\\
-4.8& 14.8\\
-1.5& -8.5\\
1.5& 8.5\\
4.8& -14.8\\
7.8 &2.2\\
10& \le0\\
\end{matrix}$$

This discussion gives seven starting intervals to refine the roots by numerical methods, such as regula falsi.
A: Firstly note we can look only on the positive part.
Secondly we can look only at $[0,10]$
Now, how many repetitions $\sin x$ do in that time? The answer is around $1.5$
Now how parts are there in the positive part? $2$
Each part goes to some value and then return to $0$
From those things we can get that we have $4$ interactions over $x\ge 0$
Multiply it by $2$ and subtract $1$(because $0$ is solution of both cases) and you get $7$
