Question

Solve: $si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}$

Answer 1

$\begin{array}{c}wehave,\\ {\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}\\ (ifx=\sin y)\\ \Rightarrow {\sin }^{-1}(1-\sin y)-2{\sin }^{-1}\left(\sin y\right)=\frac{\pi }{2}\\ \Rightarrow {\sin }^{-1}(1-\sin y)-2y=\frac{\pi }{2}\\ \Rightarrow {\sin }^{-1}(1-\sin y)=\frac{\pi }{2}+2y\\ \Rightarrow 1-\sin y=\cos 2y\\ \Rightarrow 1-\sin y=1-2{\sin }^{2}y\\ \Rightarrow 2{\sin }^{2}y=\sin y\\ \Rightarrow \sin y(2\sin y-1)=0\\ \therefore \sin y=0,x=0\\ \sin y=\frac{1}{2},x=\frac{1}{2}\\ case1:\sin y=0,x=0\\ {\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}\\ \therefore {\sin }^{-1}1=\frac{\pi }{2}\\ case2:\sin y=\frac{1}{2},x=\frac{1}{2}\\ \Rightarrow {\sin }^{-1}(1-\frac{1}{2})-2{\sin }^{-1}\frac{1}{2}=\frac{\pi }{2}\\ \Rightarrow {\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{2}=\frac{\pi }{2}\\ \therefore -{\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}\ne \frac{\pi }{2}\end{array}$