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

Question

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

Answer 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.

Answer 2

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.

Answer 3

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 $x>0, \sin x<x$: $$x^3+x^2+2x-\sin x>x^3+x^2+2x-x=x^3+x^2+x>0.$$