If $(sin^{-1}x)^2 + (cos^{-1}x)^2 =\frac{5\pi^3}{8}$, then x =

±1 $±\frac{1}{2}$ $-\frac{1}{\sqrt{2}}$ none of these

$-\frac{1}{\sqrt{2}}$

We have, $(sin^{-1}x)^2 + (cos^{-1}x)^2 =\frac{5\pi^2}{8}$ $⇒ (sin^{-1}x)^2+\left(\frac{\pi}{2}-sin^{-1}x\right)^2 = \frac{5\pi^2}{8}$ $⇒ 2(sin^{-1}x)^2 - \pi sin^{-1}x - \frac{3\pi^2}{8}=0$ $⇒  sin^{-1}x =\frac{\pi ± 2\pi}{4}⇒ sin^{-1}x = -\frac{\pi}{4}⇒x = -\frac{1}{\sqrt{2}}$