If $(sin^{-1}x)^2 + (cos^{-1}x)^2 =\frac{5\pi^2}{8}$, then x =
Answer & explanation
Correct answer: option 3
We have,
$(sin^{-1}x)^2 + (cos^{-1}x)^2 =\frac{5\pi^2}{8}$
$⇒ (sin^{-1}x)^2+\left(\frac{\pi}{2}-sin^{-1}x\right)^2 = \frac{5\pi^2}{8}$
$⇒ 2(sin^{-1}x)^2 - \pi sin^{-1}x - \frac{3\pi^2}{8}=0$
$⇒ sin^{-1}x =\frac{\pi ± 2\pi}{4}⇒ sin^{-1}x = -\frac{\pi}{4}⇒x = -\frac{1}{\sqrt{2}}$