<s>[INST] Is the equation $x^2=x \sin x+ \cos x$ true? I came across a problem that says: The equation $x^2=x \sin x+ \cos x$ is true for:  

$1.$No real value of $x$
  $2.$Exactly one  real value of $x$
  $3.$Exactly two real values of $x$
  $4.$Infinitely many real values of $x$  .

Thanks in advance for your time.
 [/INST] Consider
$f(x) = x^2-x \sin x- \cos x$.
$f(0) = -1$,
$f(\pi/2) = \pi^2/4 - \pi/2 > 0$,
and
$f(-\pi/2) = f(\pi/2) > 0$,
so $f$ has at least two real roots.
$f(-x) = f(x)$
(since $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)
= x^2 - x \sin(x) - \cos(x) = f(x)$,
 so we only need look at $x > 0$.
Since $\sin$ and $\cos$ are bounded by $1$,
$f(x) > x^2 - x - 1
> 0$
for $x > 2$ certainly
(you can get the actual root, but we only need a bound).
The only way that case (4) can hold
is if $f$ has an infinite number of roots
in a finite region
(like $\sin(1/x)$
in $(0, 1)$).
However this cannot hold
since $f'(x)$ is bounded.
So the answer seems to be (3).
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