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.
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.
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 $7$
\frac x {10}displays as $\frac x {10}$. – Dec 24 '17 at 11:08