Polynomials and Trigo unites!

$\begin{aligned}f(x)&=x^4-x^3 \sin x-x^2 \cos x\\ &=x^2(x^2-x\sin x-\cos x)\\ f(x)&=0 \implies x=0\hspace{2mm} \text { or }\hspace{2mm} x^2=xsinx+cosx\\ \\ \text{Let,}\\ g(x)&=x^2-x\sin x-\cos x \\ g(0)&=-1<0\hspace{5mm}\color{#3D99F6}\small (1)\\ g(\pi)&=\pi^2+1>0\hspace{5mm}\color{#3D99F6}\small (2)\\ g'(x)&=2x-x\cos x=x(2-\cos x)>0 \hspace{2mm} \text{for all}\hspace{2mm} x>0\\ \implies g(x)& \text{ is strictly increasing in }(0,\infty)\\ \implies g(x)& \text{has exactly one root in} (0,\infty)\hspace{5mm}\color{#3D99F6}\small \text{From }(1)\text{ and } (2)\\ g(-x)&=(-x)^2-(-x)\sin(-x)-\cos(-x)\\ &=x^2-x\sin x-\cos x\\ &=g(x)\\ \implies g(x)& \text{has exactly one root in} (0,-\infty)\\ g(x)&\text{ has 2 distinct real roots }\\ \\ \implies f(x)&\text{ has 3 real roots },0 \text { and the 2 real roots of } g(x)\end{aligned}$

$f(x)=x^{2}(x^{2} -x\sin x-\cos x)$

this will be equal to $0$ when $x=0$ or $(x^{2} -x\sin x-\cos x)=0$

for non repeated roots

$\Rightarrow \cos x= (x^{2} -x\sin x)$

the function on the right is an even continuous function.

for $x=\frac{\pi}{2}$ rhs is greater than zero but lhs is equal to zero implying we have an intersection.

First derivative of rhs suggests that it is increasing.

Combining one intersection for x>0 and one for x<0. We have two non repeated real roots. And one repeated root i.e. x = 0