Since, $\cos ^{-1} x+\sin ^{-1} x=\pi / 2$

=> $\cos ^{-1} x=\pi / 2-\sin ^{-1} x$

Substituting this in $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=17 \pi^{2} / 36$

$\left(\sin ^{-1} x\right)^{2}+\left(\pi / 2-\sin ^{-1} x\right)^{2}=17 \pi^{2} / 36$

Let $y=\sin ^{-1} x$

$\begin{array}{l}
y^{2}+((\pi / 2)-y)^{2}=17 m^{2} / 36 \\
y^{2}+\pi^{2} / 4-y^{2}-2 y((\pi / 2)-y)=17 m^{2} / 36 \\
\pi^{2} / 4-\pi y+2 y^{2}=17 \pi^{2} / 36
\end{array}$

$\begin{array}{l}
2 y^{2}-\pi y+2 / 9 m^{2}=0 \\
18 y^{2}-9 \pi y+2 m^{2}=0 \\
18 y^{2}-12 m y+3 \pi y+2 \pi^{2}=0 \\
6 y(3 y-2 \pi)+\pi(3 y-2 \pi)=0
\end{array}$

=> $(3 y-2 \pi)=0$ and $(6 y+\pi)=0$

Therefore $y=2 \pi / 3$ and $y=-\pi / 6$

substituting $y=-\pi / 6$ in $y=\sin ^{-1} x$

$\begin{array}{l}
\sin ^{-1} x=-\pi / 6 \\
x=\sin (-\pi / 6) \\
x=-1 / 2
\end{array}$

substituting $\mathrm{y}=-2 \pi / 3$ in $\mathrm{y}=\sin ^{-1} \mathrm{x}$

$\begin{array}{l}
x=\sin (2 \pi / 3) \\
x=\sqrt{3 / 2}
\end{array}$

Substituting $x=\sqrt{3} / 2$ in $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=17 \pi^{2} / 36$

$=\pi / 2$ which is not equal to $17 \pi^{2} / 36$

substituting $x=-1 / 2$ in $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=17 \pi^{2} / 36$

$\begin{array}{l}
=\pi^{2} / 36+4 \pi^{2} / 9 \\
=17 \pi^{2} / 36
\end{array}$

Hence $x=-1 / 2$