Recall that
\(\cos^2x+\sin^2x=1\)
which means
\(\sin^2x=1-\left(\dfrac5{13}\right)^2\)
\(\sin x=\pm\sqrt{\dfrac{144}{169}}=\pm\dfrac{12}{13}\)
and since \(\sin x<0\) you take the negative root:
\(\sin x=-\dfrac{12}{13}\)
Now, you know that \(\sin x<0\) for \(\pi (third and fourth quadrants). In terms of \(\dfrac x2\), you have \(\dfrac\pi2<\dfrac x2<\pi\) (second quadrant). Over this interval, you have \(\sin\dfrac x2>0\).
With this in mind, recall the half-angle identity for sine:
\(\sin^2x=\dfrac{1-\cos2x}2\)
\(\sin x=\pm\sqrt{\dfrac{1-\cos2x}2}\)
Replacing \(x\) with \(\dfrac x2\) gives
\(\sin\dfrac x2=\pm\sqrt{\dfrac{1-\cos x}2}\)
and using the fact that \(\sin\dfrac x2>0\), you take the positive root.
Finally, you have
\(\sin\dfrac x2=\sqrt{\dfrac{1-\frac5{13}}2}\)
\(\sin\dfrac x2=\dfrac2{\sqrt{13}}\)
