The number of solutions of sin⁷x + cos⁷x = 1, x∈ [0, 4π] is equal to

Solution

Maths sir · 4 steps

Compare each seventh power with the square: the sum can reach 1 only at a few special points.

1. Bound each term. Since |sin x| ≤ 1, we have sin⁷ x = sin² x · sin⁵ x ≤ sin² x. The same holds for cos x. Equality needs sin⁵ x = 1 or sin x = 0.

   sin⁷ x + cos⁷ x ≤ sin² x + cos² x = 1

2. When equality holds. The sum equals 1 only if both bounds are equalities. So sin x ∈ {0, 1} and cos x ∈ {0, 1}. That leaves (sin x, cos x) = (0, 1) or (1, 0), that is, x = 2nπ or x = 2nπ + π/2.
3. Count in the interval. In [0, 4π], x = 2nπ gives 0, 2π, 4π. x = 2nπ + π/2 gives π/2, 5π/2.
4. So. There are 3 + 2 = 5 solutions, which is option (c).

The idea to keep. For sinᵐ x + cosᵐ x = 1 with m > 2, compare with sin² x + cos² x = 1. Equality forces each of sin x and cos x to be 0 or 1. For odd m, the value -1 is not allowed.

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