Can you find a single solution of this function?

Amazing function between $cos(x)$ and $frac{sin (x)}{x}$ ! $$f(x)=sum_{n = 0}^{infty} frac{(-1)^n}{(2n+frac{1}{2})!}x^{2n}=frac{2 }{sqrt{pi }},,, _1F_2left(1;frac{3}{4},frac{5}{4};-frac{x^2}{4}right)$$

For large values of $x$, it seems to be $$f(x)simfrac{sin left(x+frac{pi }{4}right)}{sqrt{x} }-frac{1}{2 sqrt{pi } x^2}+frac{15}{8 sqrt{pi } x^4}+cdots$$ For $x=10$, the exact value is $-0.311997$ while the above truncated expansion gives $-0.311984$.

But going deeper in the simplification of the hypergeometric function $$color{red}{f(x)=sqrt {frac 2 x}left(Cleft(sqrt{frac{2x}{pi }} right) cos (x)+Sleft(sqrt{frac{2x}{pi }} right) sin (x) right)}$$

Edit

You must be very careful if you just sum the terms for a given value of $x$. For example, the partials sums $$S_p=sum_{n = 0}^{p} frac{(-1)^n}{(2n+frac{1}{2})!}10^{2n}$$ are given below to show the serious problems.

$$left( begin{array}{cc} p & S_p \ 0 & +1.12838 \ 1 & -28.9617 \ 2 & +162.087 \ 3 & -372.314 \ 4 & +465.962 \ 5 & -374.415 \ 6 & +210.195 \ 7 & -88.4566 \ 8 & +28.3181 \ 9 & -7.75129 \ 10 & +1.27170 \ 11 & -0.593517 \ 12 & -0.269554 \ 13 & -0.317495 \ 14 & -0.311378 \ 15 & -0.312058 \ 16 & -0.311992 \ 17 & -0.311997 end{array} right)$$ So, now, how many terms to be added for a given accuracy ?

Writing $$f(x)=sum_{n = 0}^{p} frac{(-1)^n}{(2n+frac{1}{2})!}x^{2n}+sum_{n = p+1}^{infty} frac{(-1)^n}{(2n+frac{1}{2})!}x^{2n}$$ we need to find $p$ such that $$frac{x^{2 (p+1)}}{left(2p+frac{5}{2}right)!} leq 10^{-k}$$ that we can rewrite as $$left(2p+frac{5}{2}right)! geq x^{2p+frac{5}{2}} frac {10^k}{sqrt x} $$

Looking at this question of mine, you will notice a superb approximation proposed by @robjohn. Applied to this case, it will give

$$color{blue}{p sim frac 12 left(x, e^{1+W(t)}-3 right)}qquad text{where}qquad color{blue}{t=frac{1}{2 e x}log left(frac{10^{2 k}}{2 pi x^2}right)}$$ Using $k=6$ and $x=10$, this gives $p=16.6868$ so $p=17$ (just as in the above table).

Notice that the exact solution would be $p=16.6872$.