Question

If ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$, then $x$ is equal to:

• A. $0,-\frac{1}{2}$
• B. $0,\frac{1}{2}$
• C. $0$
• D. None of these

Answer 1

The correct option is C

$0$

Finding the value of $x$:

Given that ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$
Let $x=\sin y$
$$\begin{gathered} \Rightarrow {\sin }^{-1}(1-\sin y)-2y=\frac{\pi }{2} \\ \Rightarrow {\sin }^{-1}(1-\sin y)=\frac{\pi }{2}+2y \\ \Rightarrow 1-\sin y=\sin \left(\frac{\pi }{2}+2y\right) \\ \Rightarrow 1-\sin y=\cos 2y \\ \Rightarrow 1-\sin y=1-2{\sin }^{2}y[\cos 2y=1-2{\sin }^{2}y] \\ \Rightarrow 2{\sin }^{2}y-\sin y=0 \\ \Rightarrow 2x^2-x=0 \\ \Rightarrow \mathrm{x}(2\mathrm{x}-1)=0 \\ \Rightarrow \mathrm{x}=0,2\mathrm{x}-1=0 \\ \Rightarrow \mathrm{x}=0,\mathrm{x}=\frac{1}{2} \end{gathered}$$
But $x=\frac{1}{2}$ does not satisfy the given equation.

So,$\mathrm{x}=0$ is the only solution to the given equation.

Hence, option C is correct.