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Solution

Let $f(x) = x^2 - x\sin x - \cos x$
Differentiating with respect to $x$:
$f'(x) = 2x - \sin x - x\cos x + \sin x$
$f'(x) = 2x - x\cos x$
$f'(x) = x(2 - \cos x)$
Since $-1 \leq \cos x \leq 1$ $\Rightarrow$ $2 - \cos x \geq 1 > 0$ always
$\therefore$ Sign of $f'(x)$ depends on sign of $x$
$f'(x) < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$
$\therefore f(x)$ has minimum at $x = 0$
Minimum value $= f(0) = 0 - 0 - \cos 0 = -1 < 0$
Also, $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)$
$= x^2 - x\sin x - \cos x = f(x)$
$\therefore f(x)$ is an even function
As $x \rightarrow \pm\infty$, $f(x) \rightarrow +\infty$
Since $f(0) = -1 < 0$ is minimum and $f(x) \rightarrow +\infty$ on both sides
$\therefore$ Graph of $f(x)$ crosses $x$-axis at exactly two points
One in $(-\infty, 0)$ and one in $(0, \infty)$
$\therefore$ Total number of points $= \boxed{2}$