Question: What are the solutions of the equation sin(2x) = 0 in the interval [0, 2π]?
Options:
0, π, 2π
0, π/2, π
0, π/4, π/2
0, 3π/2
Correct Answer: 0, π, 2π
Solution:
The solutions are x = 0, π, and 2π.
What are the solutions of the equation sin(2x) = 0 in the interval [0, 2π]?
Practice Questions
Q1
What are the solutions of the equation sin(2x) = 0 in the interval [0, 2π]?
0, π, 2π
0, π/2, π
0, π/4, π/2
0, 3π/2
Questions & Step-by-Step Solutions
What are the solutions of the equation sin(2x) = 0 in the interval [0, 2π]?
Step 1: Understand the equation sin(2x) = 0. This means we need to find the values of 2x where the sine function equals zero.
Step 2: Recall that sine equals zero at integer multiples of π. So, we can write the equation as 2x = nπ, where n is any integer.
Step 3: Solve for x by dividing both sides of the equation by 2: x = nπ/2.
Step 4: Determine the values of n that keep x within the interval [0, 2π].
Step 5: For n = 0: x = 0/2 = 0.
Step 6: For n = 1: x = 1π/2 = π/2 (not a solution).
Step 7: For n = 2: x = 2π/2 = π.
Step 8: For n = 3: x = 3π/2 (not a solution).
Step 9: For n = 4: x = 4π/2 = 2π.
Step 10: Collect the valid solutions: x = 0, π, and 2π.
Trigonometric Equations – The question tests the understanding of solving trigonometric equations, specifically using the sine function and its properties.
Periodicity of Sine Function – It assesses knowledge of the periodic nature of the sine function and how to find all solutions within a specified interval.
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