The sum of all the solution (s) of the equati | vedclass

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(C) Given equation: $\sin^{-1} 2x = \cos^{-1} x$Let $\cos^{-1} x = 	heta$,then $x = \cos 	heta$. Since the range of $\cos^{-1} x$ is $[0, \pi]$,$x$

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$\frac{1}{\sqrt{5}}$

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(C) Given equation: $\sin^{-1} 2x = \cos^{-1} x$Let $\cos^{-1} x = 	heta$,then $x = \cos 	heta$. Since the range of $\cos^{-1} x$ is $[0, \pi]$,$x$ must be in $[-1, 1]$.Also,for $\sin^{-1} 2x$ to be defined,$2x \in [-1, 1]$,so $x \in [-1/2, 1/2]$.Using the identity $\cos^{-1} x = \sin^{-1} \sqrt{1-x^2}$ for $x \ge 0$:$\sin^{-1} 2x = \sin^{-1} \sqrt{1-x^2}$$2x = \sqrt{1-x^2}$Squaring both sides: $4x^2 = 1 - x^2$$5x^2 = 1 \Rightarrow x^2 = 1/5$$x = \pm 1/\sqrt{5}$Check validity:If $x = -1/\sqrt{5}$,then $\sin^{-1}(2(-1/\sqrt{5})) = \sin^{-1}(-2/\sqrt{5})$. Since $-2/\sqrt{5} \approx -0.89$,this is a negative angle.However,$\cos^{-1}(-1/\sqrt{5})$ is in the second quadrant (positive angle).Thus,$x = -1/\sqrt{5}$ is an extraneous solution.If $x = 1/\sqrt{5}$,then $\sin^{-1}(2/\sqrt{5}) = \cos^{-1}(1/\sqrt{5})$,which is valid.The only solution is $x = 1/\sqrt{5}$.The sum of all solutions is $1/\sqrt{5}$.

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

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