Question

How do you solve `sin x * x = 0.289` ?

Answer 1

Some solutions are:

`x ~~ +-0.551533275755`

`x ~~ +-3.046589642387`

Explanation:

I will assume that you want to solve:

`(sinx)x = 0.289`

This equation has an infinite number of solutions, that is points of intersection between :

`y = sinx`

and:

`y = 0.289/x`

Here is a graph of those two functions:

graph{(y - sinx)(y - 0.289/x) = 0 [-10, 10, -5, 5]}

The smallest solutions are near `x ~~ +-0.5`

For large values of `x`, the solutions are near the zeros of `sinx`, i.e. `x ~~ kpi` for integer values of `k`.

Note that if `x` is a solution, then so is `-x` since the functions are odd.

We can find numerical approximations to the roots using Newton's method...

Let:

`f(x) = sinx - 0.289/x`

Then

`f'(x) = cosx + 0.289/x^2`

Given a first approximation `a_0` to a zero of `f(x)`, we can find successively better approximations by repeatedly applying the formula:

`a_(i+1) = a_i - f(a_i)/(f'(a_i))`

`color(white)(a_(i+1)) = a_i - (sin a_i - 0.289/a_i)/(cos a_i + 0.289/a_i^2)`

`color(white)(a_(i+1)) = a_i - (a_i^2 sin a_i - 0.289 a_i)/(a_i^2 cos a_i + 0.289)`

Putting `a_0 = 0.5` and these formulas into a spreadsheet, I got the following:

`a_0 = 0.5`

`a_1 ~~ 0.548473301868`

`a_2 ~~ 0.551522936877`

`a_3 ~~ 0.551533275637`

`a_4 ~~ 0.551533275755`

`a_5 ~~ 0.551533275755`

by changing `a_0`, approximations to the other roots are found.

For example, with `a_0 = pi`, I got:

`a_0 ~~ 3.141592653590`

`a_1 ~~ 3.046826160905`

`a_2 ~~ 3.046589645726`

`a_3 ~~ 3.046589642387`

`a_4 ~~ 3.046589642387`