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

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What’s inside this PDF?

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

Options:

  1. 0, π, 2π
  2. 0, π/3, 2π/3
  3. 0, π/2, π
  4. 0, π/4, π/2

Correct Answer: 0, π, 2π

Solution:

The solutions are x = 0, π, 2π, and x = nπ/3 for n = 0, 1, 2, 3, 4, 5.

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

Practice Questions

Q1
Solve the equation sin(3x) = 0 for x in the interval [0, 2π].
  1. 0, π, 2π
  2. 0, π/3, 2π/3
  3. 0, π/2, π
  4. 0, π/4, π/2

Questions & Step-by-Step Solutions

Solve the equation sin(3x) = 0 for x in the interval [0, 2π].
Correct Answer: x = 0, π, 2π, and x = nπ/3 for n = 0, 1, 2, 3, 4, 5.
  • Step 1: Understand the equation sin(3x) = 0. This means we need to find values of 3x 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 3x = nπ, where n is any integer.
  • Step 3: Solve for x by dividing both sides of the equation by 3: x = nπ/3.
  • Step 4: Determine the values of n that keep x within the interval [0, 2π].
  • Step 5: Calculate the values of x for n = 0, 1, 2, 3, 4, 5: This gives us x = 0, π/3, 2π/3, π, 4π/3, 5π/3, and 2π.
  • Step 6: List the solutions: x = 0, π, 2π, and x = nπ/3 for n = 0, 1, 2, 3, 4, 5.
  • Trigonometric Equations – The question tests the ability to solve trigonometric equations, specifically using the sine function and understanding its periodic nature.
  • Interval Restrictions – The question requires finding solutions within a specified interval, which is crucial for determining valid solutions.
  • Multiple Solutions – The sine function has multiple solutions within the given interval, and recognizing all possible solutions is essential.
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