Let \( \dfrac{\pi}{2} < x < \pi \) be such that \( \cot x = -\dfrac{5}{\sqrt{11}} \). Then \( \displaystyle \left(\sin \frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \frac{11x}{2}\right)(\sin 6x + \cos 6x) \) is equal to:
\( \dfrac{\sqrt{11}-1}{2\sqrt{3}} \)
\( \dfrac{\sqrt{11}+1}{2\sqrt{3}} \)
\( \dfrac{\sqrt{11}+1}{3\sqrt{2}} \)
\( \dfrac{\sqrt{11}-1}{3\sqrt{2}} \)
