Question

The sum of all the solution(s) of the equation ${\sin }^{-1}\left(2x\right)={\cos }^{-1}x$ is

• A. $0$
• B. $\frac{2}{\sqrt{5}}$
• C. $\frac{1}{\sqrt{5}}$
• D. $-\frac{1}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{5}}$

${\sin }^{-1}\left(2x\right)={\sin }^{-1}\left(\sqrt{1-x^2}\right)\Rightarrow 4x^2=1-x^2\Rightarrow x=\pm \frac{1}{\sqrt{5}}$
But only $x=\frac{1}{\sqrt{5}}$ will satisfy the condition.