How to find $x^3 - x = \sin(\pi \cdot x)$ how can I proceed to find $x$ where
$$
x^3 - x = \sin(\pi \cdot x)
$$
I'm not sure if, and how, that is possible.
Thank you.
 A: $$(1) \quad x^3 - x = \sin(\pi \cdot x)$$
Sub $\lambda=\sin(\pi \cdot x)$,
$$(2) \quad x \cdot (x^2-1)=0$$
Has solutions, $x=-1$, $\ x=0$ and $x=1$. However $\lambda=0$ at all these points. So these solutions to $(2)$ are also solutions to $(1)$.
Here's a picture,

To show that these are the only solutions, note that $x^3-x$ is monotonic on the intervals $[1,\infty)$ and $(-\infty,-1]$. And note that $\sin(\pi x)$ is monotonic on the intervals $[1,1.5]$ and $[-1.5,-1]$. Then note that derivatives of $x^3-x$ and $\sin(\pi x)$ are opposite signs on $(1,1.5)$ and $(-1.5,-1)$ and $1.5^3-1.5=1.875$ and $(-1.5)^3-(-1.5)=-1.875$. This means that there can't be any zeros on those intervals. From the there to infinity, $x^3-x$ is greater than one, which is outside the range of $\sin(\pi \cdot x)$, so there can't be any zeros there either.
So the takeaway is that you can solve both sides of an equation and then compare solutions to see what the solution of both sides of the equation put together is.
For instance, try finding solutions to,
$$x^5 \cdot (x^4-1)=\ln(x)+x-1$$
A: For $0<x<1$ we have that $\sin \pi x >0$, but $x^3-x=x(x^2-1)<0$; hence we have no root in $0<x<1$. 
Consider now $1<x<2$: here $\sin \pi x <0$ but $x^3-x=x(x^2-1)>0$; hence no root in $1<x<2$. 
Consider now $2\leq x$: here $x^3-x=x(x^2-1)\geq 6$ but $\sin \pi x \leq 1$.
Hence we should just check $x=0,1$. 
The same procedure can be followed for $x<0$.
A: Both $f(x)=x^3-x$ and $g(x)=\sin(\pi x)$ are odd functions, hence it is enough to study the non-negative solutions of $f(x)=g(x)$. $g(x)$ is a bounded function while $f(x)-g(x)$ is unbounded and increasing over $[1,+\infty)$, hence it is enough to study $h(x)=f(x)-g(x)$ over $I=[0,1]$.
Clearly $h(0)=h(1)=0$. Since $h(x)$ is a convex function on $I$, due to:
$$ h''(x) = 6x+\pi^2 \sin(\pi x) > 0, $$
$0$ and $1$ are the only roots of $h(x)$ over $[0,+\infty)$, in virtue of Rolle's theorem. It follows that $0$ and $\pm 1$ are the only roots of $h(x)$ over $\mathbb{R}$.
