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

$\forall x~(f'(x)=3x^2+2x+2+\cos x \gt 0)$ (the polynomial term is always at least $5/3$), so note that $f(-2 \pi) \lt 0, f(2 \pi) \gt 0$ and conclude that there is exactly $1$ root.
