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What is the solution to the trigonometric inequality sin(x) > cos(x) over the interval 0 ≤ x ≤ 2π radians is:

A) 0 ≤ x ≤ π/4
B) 3π/4 ≤ x ≤ π
C) π/4 ≤ x ≤ 3π/4
D) 0 ≤ x ≤ 3π/4

1 Answer

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Final answer:

The solution to the trigonometric inequality sin(x) > cos(x) over the interval 0 ≤ x ≤ 2π radians is π/4 ≤ x ≤ 3π/4. The inequality is based on the fact that sin(x) = cos(π/2 - x), and within the specified range, the cosine function decreases, making sin(x) greater than cos(x).

Step-by-step explanation:

To solve the trigonometric inequality sin(x) > cos(x) over the interval 0 ≤ x ≤ 2π radians, we first note that both sine and cosine functions range between -1 and 1. To find when sin(x) is greater than cos(x), we can look at the unit circle or use trigonometric identities.

One approach is to use the fact that sin(x) = cos(π/2 - x), which means the inequality sin(x) > cos(x) is equivalent to cos(π/2 - x) > cos(x). This inequality holds true when the angle x is in the interval (π/4, 3π/4) because in this range, the cosine function is decreasing, thus making the cos(x) value smaller than that of cos(π/2 - x).

Therefore, the correct solution for when sin(x) is greater than cos(x) in the given interval is π/4 ≤ x ≤ 3π/4, which corresponds to answer choice C).

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