The number of roots of the equation $x^3 + x^2 + 2x + \sin x = 0$ in $[-2 π, 2 π]$, is _____.

We have,

$x^3 + x^2 + 2x + \sin x = 0$   ....(i)

$⇒x^2 + x+2=-\frac{\sin x}{x},x≠0$

$⇒(x+\frac{1}{2})^2+\frac{7}{4}=-\frac{\sin x}{x}$  ....(ii)

Clearly, $x = 0$ satisfies equation (i).

For $x≠0$, we observe that

LHS >1 and 0 < RHS <-1   [$∵0<\frac{\sin x}{x}<1$ for all $x≠0$]

Hence, the given equation has just one solution.