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

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

Options:

Correct Answer: π/2

Solution:

The solution is x = π/2.

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

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

Step 1: Start with the equation 2sin(x) - 1 = 0.
Step 2: Add 1 to both sides of the equation to isolate the term with sin(x). This gives you 2sin(x) = 1.
Step 3: Divide both sides by 2 to solve for sin(x). This results in sin(x) = 1/2.
Step 4: Now, we need to find the values of x where sin(x) = 1/2 within the interval [0, 2π].
Step 5: The sine function equals 1/2 at two specific angles: x = π/6 and x = 5π/6.
Step 6: However, we are looking for the solution in the interval [0, 2π].
Step 7: The angle that corresponds to sin(x) = 1/2 in the first quadrant is x = π/6, and in the second quadrant is x = 5π/6.
Step 8: Therefore, the solutions for x in the interval [0, 2π] are x = π/6 and x = 5π/6.

Trigonometric Equations – The question tests the ability to solve a basic trigonometric equation involving the sine function.
Interval Restrictions – The solution must be found within a specified interval, which is crucial for determining valid solutions.