Question 1209759: Find the sum of solutions x ∈ R of the equation
[(3cos2x + cos³2x)/(cos⁶x - sin⁶x)] = x³ - x² + 6
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's analyze the given equation step-by-step:
**1. Simplify the Trigonometric Expression:**
* **Numerator:**
* 3cos(2x) + cos³(2x) = cos(2x)(3 + cos²(2x))
* **Denominator:**
* cos⁶(x) - sin⁶(x) = (cos²(x))³ - (sin²(x))³
* Using a³ - b³ = (a - b)(a² + ab + b²):
* (cos²(x) - sin²(x))(cos⁴(x) + cos²(x)sin²(x) + sin⁴(x))
* cos²(x) - sin²(x) = cos(2x)
* cos⁴(x) + sin⁴(x) = (cos²(x) + sin²(x))² - 2cos²(x)sin²(x) = 1 - 2cos²(x)sin²(x)
* cos⁶(x) - sin⁶(x) = cos(2x)(1 - 2cos²(x)sin²(x) + cos²(x)sin²(x))
* cos⁶(x) - sin⁶(x) = cos(2x)(1 - cos²(x)sin²(x))
* cos²(x)sin²(x) = (1/4)(2sin(x)cos(x))² = (1/4)sin²(2x)
* cos⁶(x) - sin⁶(x) = cos(2x)(1 - (1/4)sin²(2x))
* **Substitute into the Fraction:**
* [cos(2x)(3 + cos²(2x))] / [cos(2x)(1 - (1/4)sin²(2x))]
* (3 + cos²(2x)) / (1 - (1/4)sin²(2x))
* **Use cos²(2x) = 1 - sin²(2x):**
* (3 + 1 - sin²(2x)) / (1 - (1/4)sin²(2x))
* (4 - sin²(2x)) / (1 - (1/4)sin²(2x))
* Let y = sin²(2x):
* (4 - y) / (1 - y/4) = (4 - y) / ((4 - y)/4) = 4
**2. Simplify the Equation:**
* The equation becomes: 4 = x³ - x² + 6
* x³ - x² + 2 = 0
**3. Find the Roots:**
* Let's try x = -1: (-1)³ - (-1)² + 2 = -1 - 1 + 2 = 0
* Therefore, x = -1 is a root.
* We can perform polynomial division to find the other factors:
* (x³ - x² + 2) / (x + 1) = x² - 2x + 2
* The quadratic x² - 2x + 2 has discriminant (-2)² - 4(1)(2) = 4 - 8 = -4, which is negative. Therefore, it has no real roots.
**4. Find the Sum of Real Solutions:**
* The only real solution is x = -1.
**Final Answer:**
The sum of real solutions is -1.
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