Periodic functions inducing multiple roots to a (apparent)linear equations

Question

While I was solving problems,I encountered this problem:

Find number of solution to $$ \sin{x} = x/10 $$

Apparently,at first look this equation looks like a linear equation,transformable into: $$x - 10\sin{x}= 0$$

But it has about 7 roots on interval $-10 \leq x \leq 10 $

So, how could this be possible(if we go by the fundamental theorem of algebra[FTA])

Or am I overlooking something due to lack of mathematical training(Only at high -school level) because FTA is proved to be right

Is this is the tendency of periodic function? if yes then what's the reason behind it and how it's possible

Or does it happen with trigonometric functions only?If yes why?

/Why I use term periodic function instead of trigonometric?

=>I just had feeling/intuition that maybe this is generalized property of periodic functions or apply to most of the periodic functions rather than only trigonometric ones.Please feel free to correct me

Answer 1

Hint

For example let us prove there is a solution between $a=\frac{\pi}{2}$ and $b=\frac{3\pi}{2}$.

let $g(x)=sin(x)-\frac{x}{10}$.

we have

$g$ is continuous at $[a,b]$.

$g(a)=1-\frac{\pi}{20}>0$

$g(b)=-1-\frac{3\pi}{20}<0$

thus

$\exists c\in (a,b) : g(c)=0$.

you could do the same for the other intervals.

Answer 2

Transcendental functions are not applicable to the Fundamental Theorem of Algebra. Note that by invoking the FTA, you are considering $x\in\mathbb C$, upon which you will find there are not $7$, but an infinite amount of roots.

If you consider what $\sin(x)$ is as a polynomial, you may wish to see that

$$\sin(x)=x-\frac1{3!}x^3+\frac1{5!}x^5-\frac1{7!}x^7+\dots$$

where this goes on forever and $n!=1\times2\times3\times\dots\times n$. Clearly, you will get a polynomial approaching infinite degree, hence an infinite amount of roots.

This happens to most problems that include transcendental functions, which are basically functions beyond polynomials or rational functions.