Nonlinear algebraic equation with trigonometric function Does anyone have an idea how to solve following equation?
$$
(x^2 - 1)\sin(x) = 2x\cos(x)
$$
Thanks in advance!
 A: Notice analytic closed-form solutions are not possible.  First of all $x=0$ is trivial, so it doesn't count.
Secondly, we may rewrite as follows:
$$\tan(x)=\frac{2x}{x^2-1}$$
Now, if $x\ne0$ and you want a closed form solution, then it must either be trivial or a rational multiple of $\pi$.  Clearly, if $x$ is not a rational multiple of $\pi$, $\tan(x)$ is not evaluate-able unless $x=\arctan(u)$.  But if $x=\arctan(u)$ and $x$ is not a rational multiple of $\pi$, then $\frac{2x}{x^2-1}$ does not simplify!  Similarly, if $x$ is a rational multiple of $\pi$, then $\tan(x)$ is an algebraic number, but $\frac{2x}{x^2-1}$ is not an algebraic number!  See this post for more details.
So $x$ mustn't be a rational multiple of $\pi$, and we basically know nothing about its nature...except for numerical approximations!
Firstly, there are an infinite amount of solutions.  Clearly, there is at least one solution at each pole of $\tan(x)$ by intermediate value theorem.  We may also ask WolframAlpha to find these roots, but doing it ourselves is usually a far better choice of option.
Notice that
$$(x^2-1)\sin(x)=2x\cos(x)$$
$$(x^2-1)\sin(x)-2x\cos(x)=0$$
Newton's method:
$$x_{n+1}=x_n-\frac{(x_n^2-1)\sin(x_n)-2x_n\cos(x_n)}{(x_n^2-3)\cos(x_n)+4x_n\sin(x_n)}$$
with $x_0$ near any root, you can get successively better approximations.
For example, with $x_0=3.6$,
$x_1=3.6-\frac{(3.6^2-1)\sin(3.6)-2(3.6)\cos(3.6)}{(3.6^2-3)\cos(3.6)+4(3.6)\sin(3.6)}=3.67606609196$
The solution $x$ near $x_0=3.6$ can thus be approximated since $x=\lim_{n\to\infty}x_n$.
