Question

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

Answer 1

${\sin }^{-1}2x={\cos }^{-1}x\Rightarrow {\sin }^{-1}2x=\frac{\pi }{2}-{\sin }^{-1}x\Rightarrow {\sin }^{-1}2x+{\sin }^{-1}x={\sin }^{-1}1\Rightarrow {\sin }^{-1}(2x\sqrt{1-x^2}+x\sqrt{1-4x^2})={\sin }^{-1}1\Rightarrow 2x\sqrt{1-x^2}+x\sqrt{1-4x^2}=1\Rightarrow (2x\sqrt{1-x^2}{)}^2=(1-x\sqrt{1-4x^2}{)}^2\Rightarrow 4x^2(1-x^2)=1+x^2(1-4x^2)-2x\sqrt{1-4x^2}\Rightarrow 4x^2-4x^4=1+x^2-4x^4-2x\sqrt{1-4x^2}\Rightarrow 3x^2-1=-2x\sqrt{1-4x^2}\Rightarrow 9x^4+1-6x^2=4x^2(1-4x^2)\Rightarrow 9x^4+1-6x^2=4x^2-16x^4\Rightarrow 25x^4-10x^2+1=0\Rightarrow 25x^4-5x^2-5x^2+1=0\Rightarrow 5x^2(5x^2-1)-1(5x^2-1)=0(5x^2-1)(5x^2-1)=0x^2=\frac{1}{5}\Rightarrow x=\pm \frac{1}{\sqrt{5}}$

Sum of solutions $=\frac{1}{\sqrt{5}}+(-\frac{1}{\sqrt{5}})=0$