begin{aligned} & 2 sin ^{-1} x+sin ^{-1}left( | vedclass

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(B) Let $f(x) = 2 \sin ^{-1} x + \sin ^{-1}(2 x \sqrt{1-x^2}) + 3 \cos ^{-1} x - \cos ^{-1}(4 x^3 - 3 x)$.For $x \in [-1, 1]$, let $x = \cos 	heta$,

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$\pi$, when $x \in\left[-1,-\frac{1}{\sqrt{2}}\right]$

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(B) Let $f(x) = 2 \sin ^{-1} x + \sin ^{-1}(2 x \sqrt{1-x^2}) + 3 \cos ^{-1} x - \cos ^{-1}(4 x^3 - 3 x)$.For $x \in [-1, 1]$, let $x = \cos 	heta$, where $	heta \in [0, \pi]$.Then $\sin^{-1} x = \frac{\pi}{2} - 	heta$.Also, $\sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(\sin 2	heta)$ and $\cos^{-1}(4x^3-3x) = \cos^{-1}(\cos 3	heta)$.For $x \in [-1, -\frac{1}{\sqrt{2}}]$, we have $	heta \in [\frac{3\pi}{4}, \pi]$.Then $2	heta \in [\frac{3\pi}{2}, 2\pi]$, so $\sin^{-1}(\sin 2	heta) = 2	heta - 2\pi$.And $3	heta \in [\frac{9\pi}{4}, 3\pi]$, so $\cos^{-1}(\cos 3	heta) = 3\pi - 3	heta$.Substituting these: $f(x) = 2(\frac{\pi}{2} - 	heta) + (2	heta - 2\pi) + 3	heta - (3\pi - 3	heta) = \pi - 2	heta + 2	heta - 2\pi + 3	heta - 3\pi + 3	heta = 6	heta - 4\pi$.Wait, checking the range $x \in [-1, -\frac{1}{\sqrt{2}}]$ specifically for the identity $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ or $2\cos^{-1}x$:Using the standard simplification for the given expression, the value is $\pi$ for the interval $x \in [-1, -\frac{1}{\sqrt{2}}]$.

$\begin{aligned} & 2 \sin ^{-1} x+\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+3 \cos ^{-1} x \\ & -\cos ^{-1}\left(4 x^3-3 x\right) \text { is equal to }\end{aligned}$

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