For expressions of the form a sinθ + b cosθ ≥ k, combine the trigonometric functions before solving the inequality.
Example ①
Solve sinθ + cosθ ≥ 1 for 0 ≤ θ < 2π.
Answer:sinθ+cosθ = √2 sin(θ+π/4)。
√2 sin(θ+π/4) ≥ 1 → sin(θ+π/4) ≥ 1/√2。
θ+π/4 ∈ [π/4, 9π/4]。The ranges where sin ≥ √2/2 are [π/4, 3π/4] and [2π+π/4, 2π+3π/4]。
θ ∈ [0, π/2] and θ ∈ [2π, 2π+π/2] (the latter is outside the range)。
Also, θ+π/4 = 2π+π/4 → θ = 2π is not included。
0 ≤ θ ≤ π/2
Solve sinθ + cosθ ≥ 1 for 0 ≤ θ < 2π.
Answer:sinθ+cosθ = √2 sin(θ+π/4)。
√2 sin(θ+π/4) ≥ 1 → sin(θ+π/4) ≥ 1/√2。
θ+π/4 ∈ [π/4, 9π/4]。The ranges where sin ≥ √2/2 are [π/4, 3π/4] and [2π+π/4, 2π+3π/4]。
θ ∈ [0, π/2] and θ ∈ [2π, 2π+π/2] (the latter is outside the range)。
Also, θ+π/4 = 2π+π/4 → θ = 2π is not included。
0 ≤ θ ≤ π/2
Example ②
Solve √3 sinθ − cosθ ≥ 0 for 0 ≤ θ ≤ π.
Answer:2sin(θ−π/6) ≥ 0 → sin(θ−π/6) ≥ 0。
θ−π/6 ∈ [−π/6, 5π/6]。The range where sin ≥ 0 is [0, π]。The intersection is [0, 5π/6]。
θ−π/6 ∈ [0, 5π/6] → θ ∈ [π/6, π]。
π/6 ≤ θ ≤ π
Solve √3 sinθ − cosθ ≥ 0 for 0 ≤ θ ≤ π.
Answer:2sin(θ−π/6) ≥ 0 → sin(θ−π/6) ≥ 0。
θ−π/6 ∈ [−π/6, 5π/6]。The range where sin ≥ 0 is [0, π]。The intersection is [0, 5π/6]。
θ−π/6 ∈ [0, 5π/6] → θ ∈ [π/6, π]。
π/6 ≤ θ ≤ π
Key Points
- Combine → calculate the range of the new variable from the range of θ → solve the inequality → return to θ
- Intersect with the restricted range of θ to determine the final solution
Practice Problems
- Solve sinθ − cosθ ≥ 0 for 0 ≤ θ < 2π.
- Solve sinθ + √3 cosθ < 1 for 0 ≤ θ ≤ π.
- Solve 2sinθ − 2cosθ > √2 for 0 ≤ θ < 2π.
Answers and Explanations
- Answer:π/4 ≤ θ ≤ 5π/4
Explanation: √2 sin(θ−π/4) ≥ 0。θ−π/4 ∈ [−π/4, 7π/4]。sin ≥ 0 on [0,π]。θ ∈ [π/4, 5π/4]。 - Answer:0 ≤ θ < π/3 or 5π/6 < θ ≤ π
Explanation: 2sin(θ+π/3) < 1 → sin(θ+π/3) < 1/2。θ+π/3 ∈ [π/3, 4π/3] and find the range where sin < 1/2。 - Answer:5π/12 < θ < 13π/12
Explanation: 2√2 sin(θ−π/4) > √2 → sin(θ−π/4) > 1/2。θ−π/4 ∈ [−π/4,7π/4], and sin>1/2 on (π/6, 5π/6)。θ ∈ (5π/12, 13π/12)。