The sum of the solutions of the equation is trigonometry

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Solution

Q.
The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$ is [2024]

1 0

2 - 1

3 1

4 3

Ans.
(2)

We have, $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{\cos 2 x (3 + \cos^{2} 2 x)}{(\cos^{2} x - \sin^{2} x) ((\cos^{2} x + \sin^{2} x)^{2} - \cos^{2} x \sin^{2} x)} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{(3 + \cos^{2} 2 x)}{(1 - \cos^{2} x \sin^{2} x)} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{3 + {(\cos^{2} x - \sin^{2} x)}^{2}}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{3 + {(\cos^{2} x + \sin^{2} x)}^{2} - 4 \cos^{2} x \sin^{2} x}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{4 - 4 \cos^{2} x \sin^{2} x}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow 4 = x^{3} - x^{2} + 6 \Rightarrow x^{3} - x^{2} + 2 = 0$

$\Rightarrow (x + 1) (x^{2} - 2 x + 2) = 0$

$\Rightarrow x = - 1 is the only real solution.$

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