$$I = \int_{\pi/4}^{\pi/3} \left(\frac{8 \sin x - \sin 2x}{x}\right) dx$$. Then

We evaluate each inverse trigonometric term separately.

Term 1: $$\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)$$. Since $$\dfrac{2\pi}{3}$$ is not in $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, we use $$\sin\dfrac{2\pi}{3} = \sin\left(\pi - \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{3}$$. So $$\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right) = \dfrac{\pi}{3}$$.

Term 2: $$\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)$$. Since $$\dfrac{7\pi}{6}$$ is not in $$[0, \pi]$$, we use $$\cos\dfrac{7\pi}{6} = \cos\left(\pi + \dfrac{\pi}{6}\right) = -\cos\dfrac{\pi}{6}$$. Therefore $$\cos^{-1}\!\left(-\cos\dfrac{\pi}{6}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$$.

Term 3: $$\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)$$. Since $$\dfrac{3\pi}{4}$$ is not in $$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, we use $$\tan\dfrac{3\pi}{4} = \tan\left(\pi - \dfrac{\pi}{4}\right) = -\tan\dfrac{\pi}{4} = -1$$. So $$\tan^{-1}(-1) = -\dfrac{\pi}{4}$$.

Adding all three terms:

$$\dfrac{\pi}{3} + \dfrac{5\pi}{6} + \left(-\dfrac{\pi}{4}\right) = \dfrac{4\pi}{12} + \dfrac{10\pi}{12} - \dfrac{3\pi}{12} = \dfrac{11\pi}{12}$$

The answer is Option B: $$\dfrac{11\pi}{12}$$.