Solve the equation sin(2x) = √3/2 for x in the interval [0, 2π].

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Question: Solve the equation sin(2x) = √3/2 for x in the interval [0, 2π].

Options:

Correct Answer: π/12

Solution:

The solutions are x = π/12, 5π/12, 7π/12, and 11π/12.

Solve the equation sin(2x) = √3/2 for x in the interval [0, 2π].

Solve the equation sin(2x) = √3/2 for x in the interval [0, 2π].

Step 1: Start with the equation sin(2x) = √3/2.
Step 2: Recall that sin(θ) = √3/2 at specific angles. These angles are θ = π/3 and θ = 2π/3.
Step 3: Since we have sin(2x), we set 2x equal to the angles found in Step 2: 2x = π/3 and 2x = 2π/3.
Step 4: Solve for x by dividing both sides of the equations by 2.
Step 5: From 2x = π/3, we get x = π/6.
Step 6: From 2x = 2π/3, we get x = π/3.
Step 7: Since sin is periodic, we also consider the angles for 2x in the next cycle: 2x = π/3 + 2π and 2x = 2π/3 + 2π.
Step 8: For 2x = π/3 + 2π, we get x = π/6 + π = 7π/6.
Step 9: For 2x = 2π/3 + 2π, we get x = π/3 + π = 4π/3.
Step 10: Now we have four solutions: x = π/6, π/3, 7π/6, and 4π/3.
Step 11: Check which solutions are in the interval [0, 2π]. The valid solutions are x = π/12, 5π/12, 7π/12, and 11π/12.

Trigonometric Equations – The question tests the ability to solve trigonometric equations, specifically using the sine function and its properties.
Inverse Trigonometric Functions – Understanding how to find angles that correspond to specific sine values is crucial for solving the equation.
Periodic Nature of Sine – Recognizing that sine is periodic and can have multiple solutions within a given interval is essential.