Final answer:
The correct answer to the trigonometric inequality sec(2x) < -1 for 0 <= x <= 2π radians is Option C, x = 2π/3, as it falls within the range where cosine is negative and does not make sec(2x) equal to -1.
Step-by-step explanation:
To solve the trigonometric inequality sec(2x) < -1 over the interval 0 <= x <= 2π radians, we first understand that the secant function is the reciprocal of the cosine function. Therefore, sec(2x) < -1 is equivalent to cos(2x) > -1. Since the cosine function oscillates between -1 and 1 over its period, the inequality is true when the cosine function is less than 0 (since the reciprocal of a negative number is still negative).
Looking at the unit circle, cosine values are negative in the second and third quadrants. Since the angle here is 2x, we are looking for where 2x falls between π and 2π radians. To find the solutions within the given interval for x, we divide those angles by 2, so x is between π/2 and π radians, or between 3π/2 and 2π radians.
Option A (x = π/6) and Option B (x = π/2) fall out of these ranges. Option C (x = 2π/3) fits within π/2 < x < π, and Option D (x = 3π/2) fits within 3π/2 < x < 2π, which means both are part of the solution set. However, the inequality specifies sec(2x) < -1, not <= -1, so points where cos(2x) equals -1 are excluded. Cosine reaches -1 at multiples of π, thus when x is 3π/2, 2x equals 3π, at which point the cosine (and therefore secant) is exactly -1, not less than. Hence, Option D (x = 3π/2) is not part of the solution, leaving only Option C, x = 2π/3, as the correct answer.