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