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 given problem asks us to find the sum of the solutions \( x \in \mathbb{R} \) of the equation:

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

We will solve this equation step-by-step:

1. Start by simplifying the denominator \(\cos^6 x - \sin^6 x\) using the identity: \(a^6 - b^6 = (a^2 - b^2)(a^4 + a^2 b^2 + b^4)\).
2. Rewrite \(\cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x)\).
3. Further simplify \(\cos^2 x - \sin^2 x = \cos 2x\) and observe that by substituting into the numerator, \(3 \cos 2x + \cos^3 2x = \cos 2x (3 + \cos^2 2x)\).
4. The original equation is now: \(\frac{\cos 2x (3 + \cos^2 2x)}{\cos 2x P(x)} = x^3 - x^2 + 6\), where \(P(x) = \cos^4 x + \cos^2 x \sin^2 x + \sin^4 x\).
5. If \(\cos 2x \neq 0\), the \(\cos 2x\) terms cancel out, simplifying the equation to: \(\frac{3 + \cos^2 2x}{P(x)} = x^3 - x^2 + 6\).
6. However, if \(\cos 2x = 0\), then \(x = \frac{\pi}{4} + k \frac{\pi}{2}\) and observe for integer solutions: substitute to check which satisfy the polynomial. Here, only \(x = -1\) satisfies both sides being equal.
7. Thus, the equation simplifies to allow for the solution, \(x = -1\).

Conclusively, the sum of the solutions is:

• Sum: \(-1\)

Therefore, the correct answer is -1.

Answer 2

The given equation is: \(\frac{3\cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6\)

Step 1. Simplify the denominator: Using the identity \( \cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) \) and substituting \( \cos^2 x - \sin^2 x = \cos 2x \), we get:  
  \(\cos^6 x - \sin^6 x = \cos 2x \cdot (1 - \sin^2 x \cos^2 x)\)
Step 2. Rewrite the equation: Substitute this into the left side:  
  \(\frac{\cos 2x(3 + \cos^2 2x)}{\cos 2x(1 - \sin^2 x \cos^2 x)} = x^3 - x^2 + 6\)

  Simplifying further, we get:  
 \(\frac{4(3 + \cos^2 2x)}{(4 - \sin^2 2x)} = x^3 - x^2 + 6\) 
  which simplifies to:  
  \(\frac{4(3 + \cos^2 2x)}{(3 + \cos^2 2x)} = x^3 - x^2 + 6\)

Step 3. Solve the resulting polynomial equation: Expanding and rearranging terms, we get:  \(x^3 - x^2 + 2 = 0\)

  Factorizing gives:  
 \((x + 1)(x^2 - 2x + 2) = 0\)

  So the real root is \( x = -1 \).  

Step 4. Calculate the sum of real solutions: Since \( x = -1 \) is the only real solution, the sum of real solutions is:  -1
The Correct Answer is: -1