Finding number of solutions to $\sin(x)=x/10$ using an algebraic method.

Question

I am trying to find the number of solutions of the equation $\sin(x)={x/10}$. While I know about the graphical method of doing this, I want to know if there are any quicker and/or algebraic method to solve this.

Answer 1

Because the range of $\sin(x)$ is restricted to $[-1, 1]$, we only need to consider values of $x$ in the range $[-10, 10]$ since $|x/10| > 1$ when $|x| > 10$. Also, since both $\sin(x)$ and $x/10$ are odd functions, the solutions to this equation are symmetric about the $y$-axis, meaning if $x_0$ is a solution, then so is $-x_0$.

For clarity, I'm going to define $f(x) = x/10.$

First, zero is the obvious solution.

Second, since $\sin(x)$ has a slope of $1$ while $f(x)$ has a slope of $1/10$ at $x=0$, $\sin(x) > f(x)$ for small $x$. There must be a solution in the interval $[\pi/2, \pi]$ since $\sin(\pi/2) > f(\pi/2)$ and $\sin(\pi) < f(\pi)$.

From $x = \pi$ to $x = 2\pi$, there can be no solutions since $\sin(x)$ is negative here.

The function $\sin(x)$ is positive again from $x = 2\pi$ to $x = 3\pi.$ Since $3\pi < 10$, the $\sin$ function rises from $0$ to $1$ and falls from $1$ to $0$ in this interval, meaning $f(x)$ could intersect $\sin(x)$ once (tangentially) or twice. In our case, the answer is twice:

1. Once in the range $[2\pi, 5\pi/2]$ since $\sin(2\pi) < f(2\pi)$ and $\sin(5\pi/2) > f(5\pi/2)$.
2. Once in the range $[5\pi/2, 3\pi]$ since $\sin(5\pi/2) > f(5\pi/2)$ and $\sin(3\pi) < f(3\pi)$.

Each of the three positive solutions has a corresponding negative solution, so the total number of solutions is $7$.

Answer 2

This answer was made before the edit of the question

Consider that you look for the non trivial zeros of function $$f(x)=\sin(x)-\frac x {10}$$ The first derivative cancels close to $(2n+1)\frac \pi 2$ but you need to look at those which are $\in(-10,10)$.

Around these points, perform series expansions $$f(x)=\left((-1)^n-\frac{(2 n+1)}{20} \pi \right)-\frac{1}{10} \left(x-\pi \left(n+\frac{1}{2}\right)\right)-\frac{1}{2} (-1)^n \left(x-\pi \left(n+\frac{1}{2}\right)\right)^2+O\left(\left(x-\pi \left(n+\frac{1}{2}\right)\right)^4\right)$$ Solve the quadratic (keep the closest solution) to get an estimate of the solution.

For example, looking at the first non-trivial positive root $$x_{(1)}=\frac{1}{40} \left(4+60 \pi -\sqrt{3216+480 \pi }\right)=3.09411$$ while the solution, given by Newton method, is $x_{(1)}=2.85234$.

For the second non-trivial positive root $$x_{(2)}=\frac{1}{40} \left(-4-\sqrt{3216-800 \pi }+100 \pi \right)=7.09126$$ while the solution, given by Newton method, is $x_{(2)}=7.06817$.

If you want more accuracy, just use one or two iterations of Newton method. For example, the first root is poorly approximated but, using this estimate, look at Newton iterates $$\left( \begin{array}{cc} k & x_k \\ 0 & 3.0941114 \\ 1 & 2.8557329 \\ 2 & 2.8523434 \\ 3 & 2.8523419 \end{array} \right)$$