If `(sin^(-1)x)^(2)+(cos^(-1)x)^(2)=(5pi^(2))/(8)` then x =

Options

A. `pm 1`
B. `pm 1/2`
C. `-(1)/sqrt(2)`
D. none of these

Answer

D

Explanation

We have
`(sin^(-1)x)^(2)+(cos^(-1)x)^(2)=(5pi^(2))/(8)`
`rarr (sin^(-1)x)^(2)+(pi/(2)-isn^(-1)x)^(2)=(5pi^(2))/(8)`
`rarr 2(sin^(-1)x)^(2)-pi sin^(-1)x-(3pi^(2))/(8)=0`
`rarr sin^(-1)x=(pipm2pi)/(4)rarr sin^(-1)x=-(pi)/(8)=0`
`rarr sin^(-1)x=(xpm2pi)/(4)rarr sin^(-1)x=(pi)/(4)rarrx-(1)/sqrt(2)`