Using a diagram of the unit circle, find the range of angles that satisfy the inequality.
Example ①
Solve sinθ ≥ 1/2 for 0 ≤ θ < 2π.
Answer:The solutions of sinθ = 1/2 are π/6 and 5π/6. This is the range where y ≥ 1/2。
θ = π/6 ≤ θ ≤ 5π/6
Solve sinθ ≥ 1/2 for 0 ≤ θ < 2π.
Answer:The solutions of sinθ = 1/2 are π/6 and 5π/6. This is the range where y ≥ 1/2。
θ = π/6 ≤ θ ≤ 5π/6
Example ②
Solve cosθ < −√3/2 for 0 ≤ θ < 2π.
Answer:cosθ = −√3/2 → θ = 5π/6, 7π/6. This is the range where x < −√3/2。
5π/6 < θ < 7π/6
Solve cosθ < −√3/2 for 0 ≤ θ < 2π.
Answer:cosθ = −√3/2 → θ = 5π/6, 7π/6. This is the range where x < −√3/2。
5π/6 < θ < 7π/6
Key Points
- sinθ ≥ k → the arc where y ≥ k on the unit circle
- cosθ ≥ k → the arc where x ≥ k on the unit circle
- Distinguish whether endpoints are included depending on ≥ or >
Practice Problems
- Solve sinθ ≤ −√2/2 for 0 ≤ θ < 2π.
- Solve cosθ > 1/2 for 0 ≤ θ < 2π.
- Solve tanθ ≥ √3 for 0 ≤ θ < 2π.
Answers and Explanations
- Answer:5π/4 ≤ θ ≤ 7π/4
Explanation: sin = −√2/2 at 5π/4 and 7π/4. This is the lower region. - Answer:0 ≤ θ < π/3 or 5π/3 < θ < 2π
Explanation: cos=1/2 at π/3 and 5π/3. This is the right side where x > 1/2。 - Answer:π/3 ≤ θ < π/2 or 4π/3 ≤ θ < 3π/2
Explanation: tan=√3 at π/3. Within each period of π, π/3 ≤ θ < π/2。