<s>[INST] Find the number of roots of the equation, $x^3 + x^2 +2x +\sin x = 0$ in $[-2\pi , 2\pi]$. 
Find the number of roots of the equation,
$$x^3 + x^2 +2x +\sin x = 0$$
  in $[-2\pi , 2\pi]$.

What I have tried:
$$x^3 + x^2 +2x = -\sin x$$
$$x^2 +x +2 = \frac{-\sin x }{x}$$
$$(x + \frac{1}{2})^2 + \frac{7}{4} = \frac{-\sin x }{x}$$
I am getting somewhere from here, but I don't know how to continue. Please help!
The answer is 1
 [/INST] For $x<0, \sin x=x-\frac{x^3}{3!}+O(x^5)>x$:
$$x^3+x^2+2x-\sin x <x^3+x^2+2x-x=x(x^2+x+1)<0,$$
for $x>0, \sin x<x$:
$$x^3+x^2+2x-\sin x>x^3+x^2+2x-x=x^3+x^2+x>0.$$
</s>