The solution of sin^{-1} x - sin^{-1} 2x = pm | vedclass

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(D) Given equation: $\sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3}$.Taking $\sin$ on both sides:$\sin(\sin^{-1} x - \sin^{-1} 2x) = \sin(\pm \frac{\p

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$\pm \frac{1}{2}$

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(D) Given equation: $\sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3}$.Taking $\sin$ on both sides:$\sin(\sin^{-1} x - \sin^{-1} 2x) = \sin(\pm \frac{\pi}{3}) = \pm \frac{\sqrt{3}}{2}$.Using the formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$,where $A = \sin^{-1} x$ and $B = \sin^{-1} 2x$:$x \sqrt{1 - (2x)^2} - 2x \sqrt{1 - x^2} = \pm \frac{\sqrt{3}}{2}$.$x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2} = \pm \frac{\sqrt{3}}{2}$.Squaring both sides:$(x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2})^2 = \frac{3}{4}$.$x^2(1 - 4x^2) + 4x^2(1 - x^2) - 4x^2 \sqrt{(1 - 4x^2)(1 - x^2)} = \frac{3}{4}$.$x^2 - 4x^4 + 4x^2 - 4x^4 - 4x^2 \sqrt{1 - 5x^2 + 4x^4} = \frac{3}{4}$.$5x^2 - 8x^4 - \frac{3}{4} = 4x^2 \sqrt{1 - 5x^2 + 4x^4}$.Substituting $x^2 = \frac{1}{4}$:$5(\frac{1}{4}) - 8(\frac{1}{16}) - \frac{3}{4} = \frac{5}{4} - \frac{1}{2} - \frac{3}{4} = 0$.$4(\frac{1}{4}) \sqrt{1 - 5(\frac{1}{4}) + 4(\frac{1}{16})} = 1 \sqrt{1 - \frac{5}{4} + \frac{1}{4}} = 1 \sqrt{0} = 0$.Since $0 = 0$,$x = \pm \frac{1}{2}$ is the correct solution.

The solution of $\sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3}$ is

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