If sin ^{-1}(4 x)+sin ^{-1}(4 sqrt{3} x)=-fra | vedclass

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(A) Given the equation $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$.Since the range of $\sin ^{-1}(y)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}

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$\frac{1}{8}$

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(A) Given the equation $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$.Since the range of $\sin ^{-1}(y)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,for the sum to be $-\frac{\pi}{2}$,both terms must be negative or zero.Let $4x = \sin(\alpha)$ and $4\sqrt{3}x = \sin(\beta)$,where $\alpha, \beta \in [-\frac{\pi}{2}, 0]$.Then $\alpha + \beta = -\frac{\pi}{2}$,which implies $\alpha = -\frac{\pi}{2} - \beta$.Taking sine on both sides: $\sin(\alpha) = \sin(-\frac{\pi}{2} - \beta) = -\cos(\beta)$.Since $\sin^2(\beta) + \cos^2(\beta) = 1$,we have $\cos(\beta) = \sqrt{1 - \sin^2(\beta)}$.Substituting the values: $4x = -\sqrt{1 - (4\sqrt{3}x)^2}$.Squaring both sides: $16x^2 = 1 - 48x^2$.$64x^2 = 1 \implies x^2 = \frac{1}{64}$.Since $x$ must be negative for the sum to be negative,$x = -\frac{1}{8}$.The absolute value of $x$ is $|x| = \frac{1}{8}$.

If $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$,then the absolute value of $x$ is

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