Solve the equation sin(2x) = 1 for x in the interval [0, 2π].

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Question: Solve the equation sin(2x) = 1 for x in the interval [0, 2π].

Options:

Correct Answer: π/2

Solution:

The equation sin(2x) = 1 gives 2x = π/2 + 2nπ, hence x = π/4 + nπ/2. In [0, 2π], the solutions are π/4 and 5π/4.

Solve the equation sin(2x) = 1 for x in the interval [0, 2π].

Solve the equation sin(2x) = 1 for x in the interval [0, 2π].

Step 1: Start with the equation sin(2x) = 1.
Step 2: Recall that the sine function equals 1 at specific angles. The general solution for sin(θ) = 1 is θ = π/2 + 2nπ, where n is any integer.
Step 3: Set 2x equal to π/2 + 2nπ: 2x = π/2 + 2nπ.
Step 4: Solve for x by dividing everything by 2: x = π/4 + nπ.
Step 5: Now, we need to find the values of x in the interval [0, 2π].
Step 6: Start with n = 0: x = π/4.
Step 7: Next, try n = 1: x = π/4 + π = 5π/4.
Step 8: Check if there are any more solutions by trying n = 2: x = π/4 + 2π, which is greater than 2π, so we stop here.
Step 9: The solutions in the interval [0, 2π] are π/4 and 5π/4.

Trigonometric Equations – The question tests the ability to solve trigonometric equations, specifically using the sine function and its periodic properties.
Interval Notation – The question requires understanding of how to find solutions within a specified interval, [0, 2π].
Multiple Solutions – The concept of finding multiple solutions due to the periodic nature of trigonometric functions is tested.