The sum of all the solution(s) of the equatio | 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 $x \in [-\frac{1}{2}, \frac{1}{2}]$.We know $\cos^{-1} x = \sin^{-1} \sqrt{1-x^2}$ for $x \ge 0$.So,$\sin^{-1} 2x = \sin^{-1} \sqrt{1-x^2}$$\Rightarrow 2x = \sqrt{1-x^2}$Squaring both sides: $4x^2 = 1 - x^2$$\Rightarrow 5x^2 = 1$$\Rightarrow x^2 = \frac{1}{5}$$\Rightarrow x = \pm \frac{1}{\sqrt{5}}$Check the solutions:If $x = -\frac{1}{\sqrt{5}}$,then $\sin^{-1}(2(-\frac{1}{\sqrt{5}})) = \sin^{-1}(-\frac{2}{\sqrt{5}})$,which is negative. However,$\cos^{-1}(-\frac{1}{\sqrt{5}})$ is in the second quadrant (positive). Thus,$x = -\frac{1}{\sqrt{5}}$ is not a solution.If $x = \frac{1}{\sqrt{5}}$,then $\sin^{-1}(\frac{2}{\sqrt{5}}) = \cos^{-1}(\frac{1}{\sqrt{5}})$,which is true.Therefore,the only solution is $x = \frac{1}{\sqrt{5}}$.

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

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