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What are the solutions of the equation sin(2x) = 0 in the interval [0, 2π]?
What are the solutions of the equation sin(2x) = 0 in the interval [0, 2π]?
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Practice Questions
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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
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The solutions are x = 0, π, and 2π.
Questions & Step-by-step Solutions
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Q
Q: What are the solutions of the equation sin(2x) = 0 in the interval [0, 2π]?
Solution:
The solutions are x = 0, π, and 2π.
Steps: 10
Show Steps
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.
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