if

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

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

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

$\therefore(y)^{2}+\left(\frac{\pi}{2}-y\right)^{2}=\frac{17 \pi^{2}}{36}$

$\Rightarrow y^{2}+\frac{\pi^{2}}{4}+y^{2}-2 \times \frac{\pi}{2} \times y=\frac{17 \pi^{2}}{36}$

$\Rightarrow 2 y^{2}-\pi y=\frac{2 \pi^{2}}{9}$

$\Rightarrow 18 y^{2}-9 \pi y-2 \pi^{2}=0$

$\Rightarrow 18 y^{2}-12 \pi y+3 \pi y-2 \pi^{2}=0$

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

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

$\Rightarrow y=-\frac{\pi}{6} \quad\left[\right.$ Neglecting $y=\frac{2}{3} \pi$ as it is not satisfying the question]

$\therefore x=\sin y=\sin \left(-\frac{\pi}{6}\right)=-\frac{1}{2}$