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

From the comments, I knew the solution to this question. So, I am just writing it out as an answer.
For the case where $x \ne 0$,
$$(x+\frac{1}{2})^2 + \frac{7}{4} = - \frac{\sin x}{x}$$
Since, LHS is at least $\frac{7}{4}$ 
From the comment of Eric Yau, $0<\frac{\sin x}{x}<1$ and hence, $-1 < \frac{\sin x}{x} < 0$
Hence for $ x \ne 0 $, there is no solution for x.
However, clearly by substituting 0, $x=0$ is definitely a root of this equation.
Hence, $x=0$ is the only solution for this equation and thus, there is only one root for this equation.
