= Solution
<Wilson's theorem> states that an integer $n>1$ is prime if and only if $(n-1)!\equiv-1\pmod n$. For a prime $p$, every nonzero residue has a <multiplicative inverse>. Pair each residue with its inverse; the only self-inverse ones satisfy $(a-1)(a+1)=0$ in $\mathbb F_p$ and hence are $1,-1$ (the same residue when $p=2$). The resulting product is $-1$. Conversely, if $(n-1)!\equiv-1\pmod n$, no divisor $d$ of $n$ can satisfy $1<d<n$: it would divide both $(n-1)!$ and $n$, so could not divide $(n-1)!+1$. Hence $n$ is prime.
For a composite $n\ge6$ which is not the square of a prime, choose a proper factorization $n=ab$ with $1<a<b<n$ if possible. Both factors occur in $(n-1)!$, proving divisibility by $n$. If the initially chosen factors coincide but their common value is composite, choose a different proper divisor to obtain unequal factors; equivalently, the only composite integer all of whose proper nontrivial divisors coincide is a prime square. For $n=p^2\ge9$, the distinct factors $p$ and $2p$ lie below $p^2$ and their product is divisible by $p^2$. Thus $(n-1)!\equiv0\pmod n$ for every composite $n\ge6$.
For a prime $n$, write $(n-1)!=kn-1$. Then $\sin^2((n-1)!\pi/n)=\sin^2(\pi/n)$, so the summand is one. For a composite $n\ge6$ it is zero. The only remaining composite index is four, whose summand equals $\sin^2(3\pi/2)/\sin^2(\pi/4)=2$. Therefore the sum is the <prime-counting function> plus two whenever $x\ge4$; subtraction of two gives the claimed exact formula.
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