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

2

We have to find the number of solution of the equation

$x^2 = x \sin x + \cos x$

Let $f(x) = x^2$ and $g(x) = x \sin x + \cos x$

Clearly, the number of solutions of the given equation is same as the number of points of intersection of the curves $y = f (x)$ and $y = g(x)$. The curve $y = f (x)$ is a parabola with vertex at (0, 0) and y-axis as axis of symmetry.

Now, $g(x) = x \sin x + \cos x$

$⇒ g'(x) = x \cos x$

The changes in signs of $g'(x)$ are shown in the Figure.

Clearly, $x = 0,\frac{3π}{2}$ are points of relative minima and $x = ±\frac{π}{2},x=\frac{5π}{2}$ 

etc are points of relative maxima. Also, $g(x) = x \sin x + \cos x$ is an even function and $g (0) = 1, 8 (π/2)=π/2$ and $g(π)=-1$. So, a rough sketch of $y = g(x)$ is as shown in Figure.

Clearly, $y = f (x)$ and $y = g(x)$ intersect at two points.

Hence, given equation has two solutions.