If cos ^{-1} 2x + cos ^{-1} 3x = frac{pi}{3}, | vedclass

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(A) Given $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$.Using the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} (AB - \sqrt{1-A^2}\sqrt{1-B^2})$

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$\frac{\sqrt{3}}{2\sqrt{7}}$

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(A) Given $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$.Using the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} (AB - \sqrt{1-A^2}\sqrt{1-B^2})$, we get:$\cos ^{-1} (2x \cdot 3x - \sqrt{1-(2x)^2}\sqrt{1-(3x)^2}) = \frac{\pi}{3}$$6x^2 - \sqrt{1-4x^2}\sqrt{1-9x^2} = \cos \frac{\pi}{3} = \frac{1}{2}$$6x^2 - \frac{1}{2} = \sqrt{(1-4x^2)(1-9x^2)}$Squaring both sides:$(6x^2 - \frac{1}{2})^2 = (1-4x^2)(1-9x^2)$$36x^4 - 6x^2 + \frac{1}{4} = 1 - 13x^2 + 36x^4$$-6x^2 + 13x^2 = 1 - \frac{1}{4}$$7x^2 = \frac{3}{4}$$x^2 = \frac{3}{28}$$x = \sqrt{\frac{3}{28}} = \frac{\sqrt{3}}{2\sqrt{7}}$

If $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$, then $x =$

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