We solve equations of the forms sinθ = k and cosθ = k using the unit circle.
Example ①
Solve sinθ = √3/2 for 0 ≤ θ < 2π.
Answer:Reference angle π/3. sin > 0 in the first and second quadrants. θ = π/3, 2π/3
Solve sinθ = √3/2 for 0 ≤ θ < 2π.
Answer:Reference angle π/3. sin > 0 in the first and second quadrants. θ = π/3, 2π/3
Example ②
Solve 2cosθ + 1 = 0 for 0 ≤ θ < 2π.
Answer:cosθ = −1/2. Reference angle π/3. cos < 0 in the second and third quadrants. θ = 2π/3, 4π/3
Solve 2cosθ + 1 = 0 for 0 ≤ θ < 2π.
Answer:cosθ = −1/2. Reference angle π/3. cos < 0 in the second and third quadrants. θ = 2π/3, 4π/3
Example ③
Solve sin2θ = √2/2 for 0 ≤ θ < 2π.
Answer:For 0 ≤ 2θ < 4π, 2θ = π/4, 3π/4, 9π/4, 11π/4。
θ = π/8, 3π/8, 9π/8, 11π/8
Solve sin2θ = √2/2 for 0 ≤ θ < 2π.
Answer:For 0 ≤ 2θ < 4π, 2θ = π/4, 3π/4, 9π/4, 11π/4。
θ = π/8, 3π/8, 9π/8, 11π/8
Key Points
- Determine the reference angle → list the angles in each quadrant
- For equations involving kθ, multiply the range of θ by k before solving
- Solutions of tanθ = k repeat every π
Practice Problems
- Solve √2 sinθ − 1 = 0 for 0 ≤ θ < 2π.
- Solve cos2θ = 0 for 0 ≤ θ < 2π.
- Solve tanθ = −1 for 0 ≤ θ < 2π.
Answers and Explanations
- Answer:θ = π/4, 3π/4
Explanation: sinθ = √2/2. The first and second quadrants. - Answer:θ = π/4, 3π/4, 5π/4, 7π/4
Explanation: For 0 ≤ 2θ < 4π, 2θ = π/2, 3π/2, 5π/2, 7π/2。Divide by 2。 - Answer:θ = 3π/4, 7π/4
Explanation: Reference angle π/4. tan < 0 in the second and fourth quadrants.