if the

We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$

$\therefore 4 \sin ^{-1} x+\cos ^{-1} x=\pi$

$\Rightarrow 4 \sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} x=\pi \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

$\Rightarrow 3 \sin ^{-1} x=\frac{\pi}{2}$

$\Rightarrow \sin ^{-1} x=\frac{\pi}{6}$

$\Rightarrow x=\sin \frac{\pi}{6}$

$\Rightarrow x=\frac{1}{2}$

$\therefore x=\frac{1}{2}$