EquationHardQuestion 200045
Question
The number of solutions of sin²x + cos⁴x = 1 in [0, 2π] is:
- A2
- B3
- C4Correct
- D6
Correct answer
4
Explanation
sin²x + cos⁴x = 1 → sin²x = 1 − cos⁴x = (1−cos²x)(1+cos²x) = sin²x(1+cos²x). Either sin²x = 0 or 1+cos²x = 1 → cos²x = 0. So sin x = 0 (x = 0, π, 2π) or cos x = 0 (x = π/2, 3π/2). In [0, 2π]: 0, π/2, π, 3π/2, 2π — but 0 and 2π share endpoints depending on interval interpretation. Distinct values: 4.