Question

If ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\pi /2$, then x is equals ?

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

Answer 1

Answer: (C)

$\left\{0\right\}$

$si{n}^{-1}(1-x)-2si{n}^{-1}\left(x\right)=\frac{\pi }{2}\Rightarrow si{n}^{-1}(1-x)=\frac{\pi }{2}+2si{n}^{-1}\left(x\right)\Rightarrow 1-x=sin(\frac{\pi }{2}+2si{n}^{-1}(x\left)\right)\Rightarrow 1-x=cos\left(2si{n}^{-1}\right(x\left)\right)\Rightarrow si{n}^{-1}x=\alpha \Rightarrow x=sin\alpha 1-x=cos\left(2\alpha \right)\Rightarrow 1-x=1-2si{n}^2\alpha \Rightarrow 1-x=1-2x^2\Rightarrow 2x^2-x=0\Rightarrow x=0,\frac{1}{2}x=\frac{1}{2}$

Do not satisfy the equation, $x=0$