Final answer:
To solve the inequality sec(2x) < 1, we look for when the cosine function is negative since secant is the reciprocal of cosine. After analyzing the ranges in which cosine is negative, we find the solutions in the interval 0≤x≤2π radians to be x=π/2 and x=2π/3.
Step-by-step explanation:
The student's question relates to solving a trigonometric inequality: sec(2x) < 1 over the interval 0≤x≤2π radians. The secant function is the reciprocal of the cosine function. So sec(2x) < 1 is equivalent to cosine being greater than 1, which is not possible since the range of the cosine function is [-1, 1]. Therefore, we need to find where cosine is negative, as this would make the secant less than 1. The cosine function is negative in the second and third quadrants. Therefore, 2x must be in the ranges of π < 2x < 3π/2 or 3π/2 < 2x < 2π. When we solve for x, we get π/2 < x < 3π/4 or 3π/4 < x < π. This means that x = π/2 and x = 2π/3 are part of the solution set, which corresponds to options B) and C). Option D) x=3π/2 is not part of the solution set because 3π/2 is the boundary where cosine switches from negative to positive, and secant from less than 1 to greater than 1. Option A) x=π/6 is also not part of the solution set because cosine is positive in the first quadrant, which would make secant greater than 1. The correct answers to the question are B) x=π/2 and C) x=2π/3.