The sum of the solutions x in R of the equati | vedclass

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(C) Given equation: $\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6$Denominator simplification: $\cos^6 x - \sin^6 x = (\cos^2 x -

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$-1$

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(C) Given equation: $\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6$Denominator simplification: $\cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) = \cos 2x ((\cos^2 x + \sin^2 x)^2 - \cos^2 x \sin^2 x) = \cos 2x (1 - \frac{1}{4} \sin^2 2x) = \cos 2x (\frac{4 - \sin^2 2x}{4}) = \cos 2x (\frac{3 + \cos^2 2x}{4})$Substituting back into the equation: $\frac{\cos 2x (3 + \cos^2 2x)}{\cos 2x (\frac{3 + \cos^2 2x}{4})} = x^3 - x^2 + 6$Assuming $\cos 2x 
eq 0$ and $3 + \cos^2 2x 
eq 0$,we get: $4 = x^3 - x^2 + 6$$x^3 - x^2 + 2 = 0$Factoring: $(x + 1)(x^2 - 2x + 2) = 0$Since $x^2 - 2x + 2 = (x - 1)^2 + 1 > 0$ for all $x \in R$,the only real solution is $x = -1$.The sum of the real solutions is $-1$.

The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6$ is

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