Question

What is the value of x in the equation? Thank you!

`x*tan(x)-1=0`

Answer 1

There are infinite solutions, none of which has a closed form.

The first positive root is approximately `0.86033`.

Explanation:

Solutions of `xtanx-1=0` are also solutions of

`xtanx=1`

which is the same equation as

`x=1/tanx = cotx`

Thus, we are looking for all values `x` such that taking the cotangent of `x` returns `x` back (i.e. all the invariant points of `cotx`).

There does not exist a closed formula for finding invariant points of a trigonometric function. However, we can find an approximation.

Using `f(x) = xtanx-1,` its derivative `f'(x) = tanx+xsec^2 x,` and the Newton-Rhapson method

`x_(n+1)=x_n-(f(x_n))/(f'(x_n))`

for finding roots of `f(x)`, we begin with an arbitrary (but useful) estimate like `x_1=pi/4~~0.7854`:

`x_2=x_1-f(x_1)/(f'(x_1))`

`color(white)(x_2)=pi/4 -f(pi/4)/(f'(pi/4))`

`color(white)(x_2)=pi/4-(pi/4tan(pi/4)-1)/(tan(pi/4)+pi/4 sec^2 (pi/4))`

`color(white)(x_2)=pi/4-(pi/4-1)/(1+pi/4(2))`

`color(white)(x_2)=pi/4(1-(2(pi-4))/(pi(2+pi)))" " ~~ 0.868875`

A second iteration of the Newton-Rhapson method produces

`x_3=x_2-f(x_2)/(f'(x_2))`

`color(white)(x_3)~~0.868875-f(0.868875)/(f'(0.868875))`

`color(white)(x_3)~~0.868875-0.027551/(3.266703)`

`color(white)(x_3)~~0.860441`

A third produces

`x_4=x_3-f(x_3)/(f'(x_3))" "~~0.860333`

A fourth:

`x_5~~0.860334`

Our estimate of the (first positive) root of `f(x)` is now accurate to a precision of at least 5 decimal digits. Repeat the iteration procedure until the desired precision is reached.