Trigonometry Trigonometric Equations 1 — Questions and Answers
Question 1: What are all solutions to sin(x) = 0 on the interval [0, 2π)?
- x = 0, π/2
- x = 0, π (Correct answer)
- x = π/2, 3π/2
- x = π, 3π/2
Correct answer: x = 0, π
sin(x) = 0 at multiples of π, so on [0, 2π) the solutions are x = 0 and x = π.
Question 2: Solve cos(x) = 1/2 for x in [0, 2π).
- x = π/3, 5π/3 (Correct answer)
- x = π/6, 5π/6
- x = π/3, 2π/3
- x = π/4, 7π/4
Correct answer: x = π/3, 5π/3
cos(x) = 1/2 at x = π/3 (Q1) and x = 5π/3 (Q4) since cosine is positive in quadrants I and IV.
Question 3: What is the general solution for sin(x) = 1?
- x = nπ
- x = 2nπ
- x = π/2 + 2nπ (Correct answer)
- x = π/2 + nπ
Correct answer: x = π/2 + 2nπ
sin(x) = 1 only at x = π/2, and since sine has period 2π, the general solution is x = π/2 + 2nπ.
Question 4: Solve tan(x) = 1 for x in [0, 2π).
- x = π/4, 5π/4 (Correct answer)
- x = π/4, 3π/4
- x = π/6, 7π/6
- x = π/3, 4π/3
Correct answer: x = π/4, 5π/4
tan(x) = 1 at x = π/4 (Q1) and x = 5π/4 (Q3) since tangent is positive in quadrants I and III.
Question 5: Which of the following trigonometric equations has no real solution?
- sin(x) = 0.5
- cos(x) = -1
- tan(x) = 100
- sin(x) = 2 (Correct answer)
Correct answer: sin(x) = 2
The range of sin(x) is [-1, 1], so sin(x) = 2 has no real solution.
Question 6: Solve 2sin(x) − 1 = 0 for x in [0, 2π).
- x = π/6, 5π/6 (Correct answer)
- x = π/3, 2π/3
- x = π/4, 3π/4
- x = π/6, 7π/6
Correct answer: x = π/6, 5π/6
Isolating gives sin(x) = 1/2, which is satisfied at x = π/6 and x = 5π/6 in [0, 2π).
Question 7: What is the general solution for cos(x) = 0?
- x = nπ
- x = π/2 + 2nπ
- x = π/2 + nπ (Correct answer)
- x = 2nπ
Correct answer: x = π/2 + nπ
cos(x) = 0 at x = π/2 and x = 3π/2 per period, which collapses to x = π/2 + nπ for integer n.
Trigonometry Trigonometric Equations 1
What are all solutions to sin(x) = 0 on the interval [0, 2π)?