The number of solutions of the equation 4 sin ² x-4 cos ³ x+9-4 cos x=0 ; x ∈[-2 π, 2 π] is :

Solution

Maths sir · 4 steps

Turn everything into cos x, then check whether the cubic can reach 13 while cos x stays in [-1, 1].

1. One variable. Put sin² x = 1 - cos² x and write c = cos x. The equation becomes 4 - 4c² - 4c³ + 9 - 4c = 0.

   4c³ + 4c² + 4c = 13, -1 ≤ c ≤ 1

2. Range of left side. Let f(c) = 4c³ + 4c² + 4c. Then f'(c) = 12c² + 8c + 4, whose discriminant is 64 - 192 < 0, so f'(c) > 0 and f is increasing. On [-1, 1] the largest value is f(1) = 4 + 4 + 4 = 12.
3. Compare. The left side can never exceed 12, so it can never equal 13. The only real root of the cubic is about c ≈ 1.04, and cos x cannot exceed 1.
4. So. No value of x, in [-2π, 2π] or anywhere else, satisfies the equation. The number of solutions is 0, option (a).

The idea to keep. After reducing to one trig function, always check that the root lies in the allowed range. A real root of the polynomial is useless if it falls outside [-1, 1] for cos x or sin x.

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