[latex]sin \frac{ \pi*(2x-8)}{4} = \frac{ \sqrt{2}}{2} [/latex]

[latex]\sin \dfrac{ \pi(2x-8)}{4} = \dfrac{ \sqrt{2}}{2} \\ \dfrac{ \pi(2x-8)}{4}=\dfrac{\pi}{4}+2k\pi \vee \dfrac{ \pi(2x-8)}{4}=\dfrac{3\pi}{4}+2k\pi\\ \pi(2x-8)=\pi+8k\pi \vee \pi(2x-8)=3\pi+8k\pi\\ 2x-8=1+8k \vee 2x-8=3+8k\\ 2x=9+8k \vee 2x=11+8k\\ \boxed{x=\dfrac{4}{5}+4k \vee x=\dfrac{11}{2}+4k,\ (k\in\mathbb{Z})} [/latex]