1. Find the zeros of $f(x)$

For $x \neq 0$,

Since $x^2 \neq 0$ for $x\neq 0$, we have

Now, $\sin \theta =0$ when $\theta = n\pi$ for some $n\in \mathbb{Z}$. So,

Hence,

where $n\in \mathbb{N}$ (since $x^2>0$, we need $n>0$).

Also, $x=0$ is itself a zero because $f(0)=0$.

Thus, for positive $x$, the solutions are exactly

2. Check option A

Statement: $f(x)=0$ has infinitely many solutions in $\left[\frac{1}{10^{10}},\infty\right)$.

For positive zeros,

So only finitely many positive integers $n$ satisfy this. Hence only finitely many positive zeros lie in this interval.

Also, for $x>1$, there are no zeros of form $1/\sqrt{n}$ except at $x=1$. Thus the number of solutions in this interval is finite, not infinite.

So A is false.

3. Check option B

Statement: $f(x)=0$ has no solutions in $\left[\frac{1}{\pi},\infty\right)$.

We test whether some zero $x=\frac{1}{\sqrt{n}}$ lies in this interval:

Since $\pi^2 \approx 9.87$, possible integers are

So there are several solutions in this interval, e.g.

Hence B is false.

4. Check option C

Statement: The set of solutions of $f(x)=0$ in $\left(0,\frac{1}{10^{10}}\right)$ is finite.

We need

There are infinitely many integers $n>10^{20}$. Therefore, there are infinitely many zeros in this interval.

So C is false.

5. Check option D

Statement: $f(x)=0$ has more than $25$ solutions in $\left(\frac{1}{\pi^2},\frac{1}{\pi}\right)$.

We need zeros of the form $x=\frac{1}{\sqrt{n}}$ satisfying

Since all quantities are positive, invert carefully:

From

And from

Thus,

Now,

So the integers $n$ satisfying this are

Number of such integers:

Thus there are $88$ solutions in that interval, which is certainly more than $25$.

So D is true.

6. Conclusion

The only true statement is