If cos^{-1} x = alpha (0

Question

(B) Given equation is $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$.Since $\cos^{-1} x = \alpha$,we have $x = \

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$\frac{\pi}{6}$

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(B) Given equation is $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$.Since $\cos^{-1} x = \alpha$,we have $x = \cos \alpha$. Given $0 Substituting $x = \cos \alpha$ into the equation:$\sin^{-1} (2 \cos \alpha \sqrt{1 - \cos^2 \alpha}) + \sec^{-1} (\frac{1}{2 \cos^2 \alpha - 1}) = \frac{2 \pi}{3}$$\sin^{-1} (2 \cos \alpha \sin \alpha) + \sec^{-1} (\frac{1}{\cos 2 \alpha}) = \frac{2 \pi}{3}$$\sin^{-1} (\sin 2 \alpha) + \cos^{-1} (\cos 2 \alpha) = \frac{2 \pi}{3}$Since $0 Therefore,$2 \alpha + 2 \alpha = \frac{2 \pi}{3}$$4 \alpha = \frac{2 \pi}{3}$$\alpha = \frac{\pi}{6}$.

If $\cos^{-1} x = \alpha$ $(0 < x < 1)$ and $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$,then $\alpha$ is

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