Question: Solve the equation sin(2x) = 0 for x in the interval [0, 2π].
Options:
0, π, 2π
π/2, 3π/2
π/4, 3π/4
π/6, 5π/6
Correct Answer: 0, π, 2π
Solution:
The solutions are x = 0, π, 2π.
Solve the equation sin(2x) = 0 for x in the interval [0, 2π].
Practice Questions
Q1
Solve the equation sin(2x) = 0 for x in the interval [0, 2π].
0, π, 2π
π/2, 3π/2
π/4, 3π/4
π/6, 5π/6
Questions & Step-by-Step Solutions
Solve the equation sin(2x) = 0 for x in the interval [0, 2π].
Step 1: Start with the equation sin(2x) = 0.
Step 2: Recall that the sine function equals zero at integer multiples of π. So, we set 2x = nπ, where n is an integer.
Step 3: Solve for x by dividing both sides 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; for n = 1, x = π/2; for n = 2, x = π; for n = 3, x = 3π/2; for n = 4, x = 2π.
Step 6: List the valid solutions: x = 0, π, 2π.
Trigonometric Equations – The question tests the ability to solve equations involving trigonometric functions, specifically the sine function.
Periodicity of Sine Function – Understanding the periodic nature of the sine function is crucial for finding all solutions within a specified interval.
Interval Notation – The question requires knowledge of how to interpret and work within a given interval, [0, 2π].
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