Handy way to find the $x$ value where $\sin x \cos \left( \frac{\pi}{2} \sin x \right)$ is maximum? Like in the title, is there a handy way to compute the $x$ values for which the function
$$f(x) = \sin x \cos \left( \frac{\pi}{2} \sin x \right)$$
reaches its maxima?
The derivative is
$$f'(x) = \cos (x) \cos \left( \frac{\pi}{2} \sin x \right) - \sin(x) \sin \left( \frac{\pi}{2} \sin x \right) \frac{\pi}{2} \cos (x)$$
I can put $f'(x) = 0$ but I don't think it is a quick way to find the maxima's locations.
Are there some observation that can be done in order to quickly find those values of $x$?

I can't completely answer your question, but here's what I've got so far:
$f'(x) = \cos x \cos(\frac{\pi}{2} \sin x) - \frac{\pi}{2}\sin x \cos x \sin(\frac{\pi}{2} \sin x)$ $= \cos x(\cos(\frac{\pi}{2} \sin x) - \frac{\pi}{2}\sin x \sin(\frac{\pi}{2} \sin x))$
Setting this equal to $0$, we either have $\cos x = 0$ or $\cos(\frac{\pi}{2} \sin x) = \frac{\pi}{2} \sin x \sin (\frac{\pi}{2} \sin x)$. In the former case, $\sin x = \pm 1$, so $f(x) = 0$. This obviously does not correspond to the maximum of $f(x)$.
In the latter case, we need to solve $1 = \frac{\pi}{2} \sin x \tan(\frac{\pi}{2} \sin x)$. We can make our lives easier by setting $u = \frac{\pi}{2} \sin x$, so now we're solving $$1 = u \tan u$$ I am not aware of any easy solutions to this equation. More than likely, there are no elementary solutions. To proceed, you'll need to use numerical methods.
