Number of roots of $x^4 + x^3 \sin(x) + x^2 \cos(x) = 0$

Question

How many roots do the following polynomial have?

$$x^4 + x^3 \sin(x) + x^2 \cos(x)$$

Obviously $0$ is a root. Dividing by $x^2$, I am left with

$$x^2 + x\sin(x) + \cos(x)$$

How do I know this polynomial has no root?

Answer 1

Look for extremum of $f(x)=x^2 + x\sin(x) + \cos(x)$ $$ \frac{d}{dx}(x^2 + x\sin(x) + \cos(x))=2x+x\cos(x)=x(2+\cos(x))=0 $$ Then the only extremum we have is at $x=0$

We know that $x(2+\cos(x))<0$ for $x<0$ and $x(2+\cos(x))>0$ for $x>0$, also $f(0)=1$, which means that $f(x)\ge1$ for all $x$, hence it have no roots.

Answer 2

Another way is as follows: the function $$f(x)= x^4 + x^3 \sin(x) + x^2 \cos(x)$$ is an even function (since the cosinus is even and the sinus is odd), hence it is enough to see at $x\ge 0$. The derivative $$f’(x)=(4+\cos(x))x^3+2x^2\sin(x) +2x\cos(x)$$ is clearly positive for $x$ positive. Consequently $f(x)$ is increasing for $x\gt 0$. Thus the only root is the obvious $x=0$.