Question

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

Answer 1

$$\begin{gathered} 4{\sin }^{-1}x=\mathrm{\pi }-{\cos }^{-1}x \\ \Rightarrow 4{\sin }^{-1}x=\mathrm{\pi }-\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\mathrm{x}\right)\left[\because {\cos }^{-1}x=\frac{\mathrm{\pi }}{2}-{\sin }^{-1}x\right] \\ \Rightarrow 4{\sin }^{-1}x=\frac{\mathrm{\pi }}{2}+{\sin }^{-1}x \\ \Rightarrow 3{\sin }^{-1}x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\sin }^{-1}x=\frac{\mathrm{\pi }}{6} \\ \Rightarrow \mathrm{x}=\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2} \end{gathered}$$