Implications of $X( -j\omega ) = X^*(j\omega)$
The fact is that if $x(t)$ is real and then $X(-j\omega) = X^*(j\omega)$ (it's easy to prove) and, because of duality, if $X(j\omega)$ is real, so also then $x(-t) = x^*(t)$.  in your case, you have both $x(t)$ and $X(j\omega)$ being real.  keep in mind that anything that is purely real is also equal to its complex conjugate.

so you can say:

"if $x(t)$ is real and even (that is $x(-t) = x(t)$) so also is $X(j\omega)$ real and even ($X(-j\omega) = X(j\omega)$)."