Question

If ${\sin }^{-1}(1-x)+2{\sin }^{-1}\left(x\right)=\frac{\pi }{2}$, then $x$ equals

• A. $\left\{0,-\frac{1}{2}\right\}$
• B. $\left\{\frac{1}{2},0\right\}$
• C. $\left\{0\right\}$
• D. $\left\{-1,0\right\}$

Answer 1

The correct option is B

$\left\{\frac{1}{2},0\right\}$

Explanation for the correct option.

Step 1: Solve the given equation

Given that, ${\sin }^{-1}(1-x)+2{\sin }^{-1}\left(x\right)=\frac{\pi }{2}$

$$\begin{gathered} \Rightarrow {\sin }^{-1}(1-x)=\frac{\mathrm{\pi }}{2}-2{\sin }^{-1}\left(x\right) \\ \Rightarrow 1-x=\sin \left[\frac{\mathrm{\pi }}{2}-2{\sin }^{-1}\left(x\right)\right] \\ \Rightarrow 1-x=\cos \left[2{\sin }^{-1}\left(x\right)\right]\dots .\text{(i)} \end{gathered}$$

Put $2{\sin }^{-1}\left(x\right)=A$
$\begin{array}{rcl}\Rightarrow {\sin }^{-1}\left(x\right)& =& \frac{A}{2}\\ & \Rightarrow & \sin \frac{A}{2}=x\end{array}$

Step 2: Find the value of $x$
Using the identity $\cos 2\mathrm{A}=1-2{\sin }^2\mathrm{A}$
$\begin{array}{rcl}& \Rightarrow & \cos A=1-2{\sin }^2\left(\frac{A}{2}\right)\\ & \Rightarrow & \cos A=1-2x^2\\ & \Rightarrow & A={\cos }^{-1}\left(1-2x^2\right)\end{array}$
From (i):
$$\begin{gathered} \Rightarrow 1-x=\cos \left[{\cos }^{-1}\left(1-2x^2\right)\right] \\ \Rightarrow 1-x=1-2x^2 \\ \Rightarrow 2x^2-x=0 \\ \Rightarrow x(2x-1)=0 \\ x=0,2x-1=0 \\ x=0,\frac{1}{2} \end{gathered}$$

Hence, option B is correct.