Question

The sum of the solutions \( x \in \mathbb{R} \) of the equation\[\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6\]is

• A. 0
• B. 1
• C. -1
• D. 3

Answer 1

The Correct Option is C

The objective is to determine the sum of real solutions \( x \) for the equation:

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

The equation is solved sequentially:

1. The denominator, \( \cos^6 x - \sin^6 x \), is simplified using the algebraic identity \( a^6 - b^6 = (a^2 - b^2)(a^4 + a^2 b^2 + b^4) \).
2. This yields \( \cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) \).
3. Recognizing \( \cos^2 x - \sin^2 x = \cos 2x \) and factoring the numerator as \( \cos 2x (3 + \cos^2 2x) \) is performed.
4. The equation transforms to \( \frac{\cos 2x (3 + \cos^2 2x)}{\cos 2x P(x)} = x^3 - x^2 + 6 \), where \( P(x) = \cos^4 x + \cos^2 x \sin^2 x + \sin^4 x \).
5. Assuming \( \cos 2x eq 0 \), the equation simplifies to \( \frac{3 + \cos^2 2x}{P(x)} = x^3 - x^2 + 6 \) after cancellation.
6. When \( \cos 2x = 0 \), which occurs at \( x = \frac{\pi}{4} + k \frac{\pi}{2} \) for integer \( k \), we check for integer solutions. The value \( x = -1 \) satisfies the equality on both sides.
7. Thus, \( x = -1 \) is established as a solution.

In conclusion, the sum of the solutions is:

• Sum: \(-1\)

The final answer is -1.