Let S₁ = { x ∈ [0,12π ]:sin ⁵x + cos ⁵x = 1}
and S₂ = { x ∈ [0,8π ]:sin ⁷x + cos ⁷x = 1}
Then n(S₁) - n(S₂) is equal to ______________.

Solution

Maths sir · 4 steps

Compare each odd power with the square: sin⁵ x ≤ sin² x, so the equation is tight only at 0 and 1.

1. Bound each term. Since |sin x| ≤ 1, we get sin⁵ x ≤ |sin x|⁵ ≤ sin² x. In the same way, cos⁵ x ≤ cos² x.

   sin⁵ x + cos⁵ x ≤ sin² x + cos² x = 1

2. When equality holds. The sum equals 1 only if both inequalities are equalities. So sin⁵ x = sin² x and cos⁵ x = cos² x, which means each of sin x and cos x is 0 or 1. The same argument works for the power 7.
3. The solutions. The only pairs are (sin x, cos x) = (0, 1) or (1, 0). That gives x = 2nπ or x = π/2 + 2nπ. The value -1 is ruled out because a negative odd power only lowers the sum.

   Set	x = 2nπ	x = π/2 + 2nπ	Total
   S₁, [0, 12π]	n = 0 to 6: 7 values	n = 0 to 5: 6 values	13
   S₂, [0, 8π]	n = 0 to 4: 5 values	n = 0 to 3: 4 values	9

4. So. n(S₁) - n(S₂) = 13 - 9 = 4.

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 never works.

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