Finding real solutions to $\sin { x=\frac { x }{ 50 } }$ Here's a question that came to my mind,

Find the number of real valued solutions to the equation $\sin { x=\frac { x }{ 50 }  }$.

I am confused and not able to move on. Please help.
Thank you.
 A: Solutions are in $[-50,50].$ As $\pi\cdot15\simeq47.12,$ using a table of variations for $\sin x-\frac{x}{50}$ (which is an odd function), you can see that there is $15\cdot 2+\underset{\text{the $0$ one}}{\underbrace{1}}=31$ solutions, but you can't get them as real numbers.
A: We can solve this problem using a Taylor series expansion to get approximate values:$$\sin(x)=\frac x{50}$$$$\sin(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{(2n-1)!}=\frac x{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$$$$\frac x{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots=\frac x{50}$$The trick here is to do approximations, not the actual thing.$$\frac{x}{1!}=\frac x{50}\to x_1=0$$$$\frac x{1!}-\frac{x^3}{3!}=\frac{x}{50}\to x_2=0,\pm\frac{7\sqrt{3}}{5}$$Repeating the process, we will get an infinite amount of answers, assuming we allow complex answers.
We will also note $x=0$ will always be a solution, one we can see graphically and within any of the above approximations we want.
A: If you see the graph of it you can see the fluctuations and meet  $15$  times with $X-axis$ before $π$ on one side. now range of $x$ can be $[-50,50]$ so total solutions are $15+15+1=31$. $+1$ as $0$ is also one of the root.
