The number of points in $\left(-\infty ,\infty \right)$, for which $x^2-x\sin x-\cos x=0$, is:

$f\left(x\right)=x^2-x\sin x-\cos x$

$f^{\text{'}}\left(x\right)=2x-x\cos x-\sin x+\sin x$

 $=x\left(2-\cos x\right)$

We know that, $\left(2-\cos x\right)$ is positive for all real numbers, now it depends on $x$. 

So,$f\left(x\right)$is increasing if $x>0$

$f\left(x\right)$ is decreasing if $x<0$

$f\left(0\right)=-1$

$f\left(\infty \right)=\infty$

$f\left(-\infty \right)=\infty$

So, it is clear that graph cuts the $x$-axis at two points, so number of solutions are two.