The sum of the solutions x ∈ ℝ of the equation (3 cos 2 x+cos ³ 2 x)/(cos ⁶ x-sin ⁶ x)=x³-x²+6 is

Solution

Maths sir · 4 steps

The left side looks scary, but it collapses to the constant 4, so you only need to solve a cubic.

1. Factor the denominator. Use a³ - b³ with a = cos² x, b = sin² x. The first factor is cos² x - sin² x = cos 2x. The second is cos⁴ x + cos² x sin² x + sin⁴ x = 1 - sin² x cos² x = 1 - (sin² 2x)/4.
2. Match the numerator. Since sin² 2x = 1 - cos² 2x, the second factor is (3 + cos² 2x)/4. The numerator is cos 2x (3 + cos² 2x), so everything cancels except the 4, as long as cos 2x ≠ 0.

   (cos 2x (3+cos² 2x))/(cos 2x · (3+cos² 2x)/4) = 4

3. Solve the cubic. Now x³ - x² + 6 = 4, that is x³ - x² + 2 = 0. Here x = -1 is a root, and x³ - x² + 2 = (x+1)(x² - 2x + 2). The quadratic has discriminant 4 - 8 = -4 < 0, so it has no real roots.
4. So. The only real solution is x = -1, and cos(-2) ≠ 0, so it is allowed. The sum of the real solutions is -1, option (d).

The idea to keep. When a trig expression is set equal to a polynomial in x, first simplify the trig side. It often becomes a constant, and the problem turns into plain algebra. Then keep only the real roots.

Answer from the NTA final answer key. Solution by PrepWiser.