If cos ^{-1} x = sin ^{-1}(3x),then x is equa | vedclass

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(A) Given,$\cos ^{-1} x = \sin ^{-1} 3x$.We know that $\cos ^{-1} x = \sin ^{-1} \sqrt{1-x^2}$ for $x \in [0, 1]$.Substituting this into the equation,

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$\frac{\sqrt{10}}{10}$

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(A) Given,$\cos ^{-1} x = \sin ^{-1} 3x$.We know that $\cos ^{-1} x = \sin ^{-1} \sqrt{1-x^2}$ for $x \in [0, 1]$.Substituting this into the equation,we get $\sin ^{-1} \sqrt{1-x^2} = \sin ^{-1} 3x$.Taking sine on both sides,we have $\sqrt{1-x^2} = 3x$.Squaring both sides,we get $1-x^2 = 9x^2$.This simplifies to $10x^2 = 1$,which gives $x^2 = \frac{1}{10}$.Since $x$ must be positive for the domain of $\sin ^{-1} 3x$ and $\cos ^{-1} x$,we take $x = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$.

If $\cos ^{-1} x = \sin ^{-1}(3x)$,then $x$ is equal to

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