Da equação do enunciado, temos que: $(\underbrace{\sin^2x + \cos^2x}_1 )^2 \pm 2\sin^2x \cos^2x$ $=$ $\large{\frac{5}{8}}$ , logo: $\large{\frac{3}{8}} = \frac{\sin^2(2x)}{2}$ , assim: $\sin(2x) = \pm \large{\frac{\sqrt3}{2}}$ $\implies$ $x = \large{\{\frac{\pi}{6}, \space\frac{\pi}{3}, \space\frac{2\pi}{3}, \space\frac{5\pi}{6} \}}$ , os inversos de $x$ são: $\large{\{\frac{6}{\pi}, \space\frac{3}{\pi}, \space\frac{3}{2\pi}, \space\frac{6}{5\pi} \}}$, a soma então é: $\boxed{\large{\frac{6}{\pi}+\frac{3}{\pi}+\frac{3}{2\pi}+\frac{6}{5\pi} } = \frac{117\pi}{10}}$ Alternativa $\mathbb{(B)}$