Lesson 3: Equations Involving Multiple Angles
Estimated time: 30-35 minutes
Learning Objectives
- Solve equations of the form sin(nx) = k or cos(nx) = k
- Adjust the interval to account for the multiple angle
- Find all solutions within the original interval
- Handle equations with half-angles
Strategy for Multiple Angles
To solve sin(nx) = k on [0, 2π): First solve for nx on [0, 2nπ), then divide all solutions by n to get x values in [0, 2π).
The key insight: if the angle is nx and we want x in [0, 2π), then nx ranges over [0, 2nπ). So we need to find more solutions than usual before dividing.
Worked Examples
Example 1: sin(2x) = √3/2 on [0, 2π)
Let u = 2x. We need u in [0, 4π).
sin u = √3/2 ⇒ u = π/3, 2π/3 (in first period) and u = π/3 + 2π, 2π/3 + 2π (in second period)
u = π/3, 2π/3, 7π/3, 8π/3
x = u/2: x = π/6, π/3, 7π/6, 4π/3
Example 2: cos(3x) = 0 on [0, 2π)
Let u = 3x. u ranges over [0, 6π).
cos u = 0 ⇒ u = π/2, 3π/2, 5π/2, 7π/2, 9π/2, 11π/2
x = u/3: x = π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6
Example 3: tan(x/2) = 1 on [0, 2π)
Let u = x/2. u ranges over [0, π).
tan u = 1 ⇒ u = π/4 (only solution in [0, π))
x = 2u = π/2
Check Your Understanding
1. Solve cos(2x) = 1/2 on [0, 2π).
2. How many solutions does sin(4x) = 0 have on [0, 2π)?
3. Solve tan(2x) = −1 on [0, 2π).
Key Takeaways
- For sin(nx)=k, expand the interval for the substitution variable to [0, 2nπ).
- Find all solutions of the substitution, then divide by n.
- Multiple-angle equations usually have more solutions than basic equations.