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Mathematics Question on Differential Equations

The sum of the solutions x∈Rx \in \mathbb{R} of the equation3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6is

A

0

B

1

C

-1

D

3

Answer

-1

Explanation

Solution

The given equation is: 3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\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⁡6x−sin⁡6x=(cos⁡2x−sin⁡2x)(cos⁡4x+cos⁡2xsin⁡2x+sin⁡4x)\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⁡2x−sin⁡2x=cos⁡2x\cos^2 x - \sin^2 x = \cos 2x, we get:
cos⁡6x−sin⁡6x=cos⁡2x⋅(1−sin⁡2xcos⁡2x)\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:
cos⁡2x(3+cos⁡22x)cos⁡2x(1−sin⁡2xcos⁡2x)=x3−x2+6\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:
4(3+cos⁡22x)(4−sin⁡22x)=x3−x2+6\frac{4(3 + \cos^2 2x)}{(4 - \sin^2 2x)} = x^3 - x^2 + 6
which simplifies to:
4(3+cos⁡22x)(3+cos⁡22x)=x3−x2+6\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: x3−x2+2=0x^3 - x^2 + 2 = 0

Factorizing gives:
(x+1)(x2−2x+2)=0(x + 1)(x^2 - 2x + 2) = 0

So the real root is x=−1x = -1.

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