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 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