Let cos ^{-1}(x) + cos ^{-1} (2x) + cos ^{-1} | vedclass

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(C) Given $\cos ^{-1}(x) + \cos ^{-1}(2x) + \cos ^{-1}(3x) = \pi.$Rearranging the terms,we get $\cos ^{-1}(2x) + \cos ^{-1}(3x) = \pi - \cos ^{-1}(x).

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$26$

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(C) Given $\cos ^{-1}(x) + \cos ^{-1}(2x) + \cos ^{-1}(3x) = \pi.$Rearranging the terms,we get $\cos ^{-1}(2x) + \cos ^{-1}(3x) = \pi - \cos ^{-1}(x).$Using the identity $\cos ^{-1}(A) + \cos ^{-1}(B) = \cos ^{-1}(AB - \sqrt{1-A^2}\sqrt{1-B^2})$ and $\pi - \cos ^{-1}(x) = \cos ^{-1}(-x),$$\cos ^{-1}(6x^2 - \sqrt{1-4x^2}\sqrt{1-9x^2}) = \cos ^{-1}(-x).$Equating the arguments,$6x^2 - \sqrt{1-4x^2}\sqrt{1-9x^2} = -x.$Rearranging,$6x^2 + x = \sqrt{1-4x^2}\sqrt{1-9x^2}.$Squaring both sides,$(6x^2 + x)^2 = (1-4x^2)(1-9x^2).$$36x^4 + x^2 + 12x^3 = 1 - 13x^2 + 36x^4.$Simplifying,$12x^3 + 14x^2 - 1 = 0.$Comparing this with $ax^3 + bx^2 + cx - 1 = 0,$ we get $a = 12, b = 14, c = 0.$Thus,$a + b + c = 12 + 14 + 0 = 26.$

Let $\cos ^{-1}(x) + \cos ^{-1} (2x) + \cos ^{-1}(3x) = \pi.$ If $x$ satisfies the cubic equation $ax^3 + bx^2 + cx - 1 = 0,$ then the value of $(a + b + c)$ is -

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