The equation cos ^{-1}(1-x)-2 cos ^{-1} x=fra | vedclass

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(B) Given equation: $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$.Let $\cos ^{-1} x = 	heta$,then $x = \cos 	heta$,where $	heta \in [0, \pi]$.The

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(B) Given equation: $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$.Let $\cos ^{-1} x = 	heta$,then $x = \cos 	heta$,where $	heta \in [0, \pi]$.The equation becomes $\cos ^{-1}(1-\cos 	heta) - 2	heta = \frac{\pi}{2}$.$\cos ^{-1}(1-\cos 	heta) = \frac{\pi}{2} + 2	heta$.Taking $\cos$ on both sides:$1-\cos 	heta = \cos(\frac{\pi}{2} + 2	heta) = -\sin(2	heta)$.$1-\cos 	heta = -2\sin 	heta \cos 	heta$.$1 - (1-2\sin^2(\frac{	heta}{2})) = -2(2\sin(\frac{	heta}{2})\cos(\frac{	heta}{2}))\cos 	heta$.$2\sin^2(\frac{	heta}{2}) = -4\sin(\frac{	heta}{2})\cos(\frac{	heta}{2})\cos 	heta$.Case $1$: $\sin(\frac{	heta}{2}) = 0 \implies 	heta = 0 \implies x = \cos 0 = 1$.Check $x=1$: $\cos ^{-1}(0) - 2\cos ^{-1}(1) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. This is a solution.Case $2$: $\sin(\frac{	heta}{2}) = -2\cos(\frac{	heta}{2})\cos 	heta$.$	an(\frac{	heta}{2}) = -2\cos 	heta$. Since $	heta \in [0, \pi]$,$	an(\frac{	heta}{2}) \ge 0$ and $-2\cos 	heta$ can be negative. For $	heta \in [0, \pi/2]$,$\cos 	heta \ge 0$,so $-2\cos 	heta \le 0$. The only intersection is at $	heta=0$ (already found).Thus,there is only one solution.

The equation $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$ has

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