If k = tan(frac{pi}{4} + frac{1}{2}cos^{-1}(f | vedclass

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(A) Let $\alpha = \frac{1}{2}\cos^{-1}(\frac{2}{3})$ and $\beta = \frac{1}{2}\sin^{-1}(\frac{2}{3})$.Then $k = 	an(\frac{\pi}{4} + \alpha) + 	an(\be

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(A) Let $\alpha = \frac{1}{2}\cos^{-1}(\frac{2}{3})$ and $\beta = \frac{1}{2}\sin^{-1}(\frac{2}{3})$.Then $k = 	an(\frac{\pi}{4} + \alpha) + 	an(\beta)$.Since $2\alpha = \cos^{-1}(\frac{2}{3})$, $\cos(2\alpha) = \frac{2}{3}$.Since $2\beta = \sin^{-1}(\frac{2}{3})$, $\sin(2\beta) = \frac{2}{3}$.Note that $\beta = \frac{\pi}{4} - \alpha$ is not necessarily true, but $	an(\frac{\pi}{4} + \alpha) = \frac{1 + 	an \alpha}{1 - 	an \alpha}$.Using $	an \alpha = \sqrt{\frac{1-\cos 2\alpha}{1+\cos 2\alpha}} = \sqrt{\frac{1-2/3}{1+2/3}} = \sqrt{\frac{1/3}{5/3}} = \frac{1}{\sqrt{5}}$.Using $	an \beta = \sqrt{\frac{1-\cos 2\beta}{1+\cos 2\beta}} = \sqrt{\frac{1-\sqrt{1-4/9}}{1+\sqrt{1-4/9}}} = \sqrt{\frac{1-\sqrt{5}/3}{1+\sqrt{5}/3}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} = \frac{3-\sqrt{5}}{2}$.Calculating $k$ leads to $k=3$.The equation becomes $\sin^{-1}(3x-1) = \sin^{-1}x - (\frac{\pi}{2} - \sin^{-1}x) = 2\sin^{-1}x - \frac{\pi}{2}$.Taking $\sin$ on both sides: $3x-1 = \sin(2\sin^{-1}x - \frac{\pi}{2}) = -\cos(2\sin^{-1}x) = -(1-2x^2) = 2x^2-1$.$2x^2 - 3x = 0 \Rightarrow x(2x-3) = 0$.So $x=0$ or $x=1.5$.Checking $x=1.5$: $\sin^{-1}(3.5)$ is undefined.Checking $x=0$: $\sin^{-1}(-1) = -\frac{\pi}{2}$ and $\sin^{-1}(0) - \cos^{-1}(0) = 0 - \frac{\pi}{2} = -\frac{\pi}{2}$.Thus, $x=0$ is the only solution.

If $k = \tan(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}(\frac{2}{3})) + \tan(\frac{1}{2}\sin^{-1}(\frac{2}{3}))$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is . . . . . . .

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