Solve the trigonometric equation $2 \alpha + \sin(\alpha \pi) - \alpha \cos(\alpha \pi)=0$ How we can solve in $\mathbb{R}$ the following equation?
$$
2 \alpha + \sin(\alpha \pi) - \alpha \cos(\alpha \pi)=0
$$
I'm lost.

Define
$$
f(x)=2x+\sin (\pi x)-x\cos (\pi x),
$$
then $f(0)=0$. Now we will show that $x=0$ is the unique root of $f(x)=0$. For $x> 1$,
$$
2x+\sin (\pi x)-x\cos (\pi x)\ge 2x-1-x=x-1>0
$$
and $f(1)=3>0$.
Also, in $(1/2,1)$,
$$
2x+\sin (\pi x)-x\cos (\pi x) \ge 1+\sin (\pi x) -1 =\sin (\pi x)>0
$$
and $f(1/2)=2>0$.
Find derivative of $f(x)$:
$$
f'(x)=2+(\pi-1)\cos(\pi x)+\pi x \sin (\pi x)
$$
In $(0,1/2)$, $f'(x)>0$ and $f(0)=0$, so $f(x)=0$ has no root in $(0,1/2)$.
Therefore, $f(x)=0$ has no root in $(0,\infty)$. Since $f(x)$ is odd, $f(x)=0$ has no root in $(-\infty,0)$, too.
Here is the graph of $y=f(x)$:

Edit: I found an easier solution, which don't need differentiation.
$$
f(x)=x(2-\cos (\pi x))+\sin (\pi x)>0
$$
for $0<x<1$, because $2-\cos (\pi x)>0$ and $\sin(\pi x)>0$. Also, we already showed $f(x)>0$ for $x\ge 1$. Using the fact that $f(x)$ is odd, we get the same conclusion.
