Question

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

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

Answer 1

The correct option is A $0$

${\sin }^{-1}2x={\cos }^{-1}x$

$\Rightarrow {\sin }^{-1}2x={\sin }^{-1}\sqrt{1-x^2}$

$\Rightarrow 2x=\sqrt{1-x^2}$

$\Rightarrow {4x}^2=1-x^2\Rightarrow {5x}^2=1$

$\Rightarrow x^2=\frac{1}{5}\Rightarrow x=\pm \frac{1}{\sqrt{5}}$

$\therefore$ Sum of all solutions $=\frac{1}{\sqrt{5}}+(-\frac{1}{\sqrt{5}})=0$ [A]