Solving Trigonometric Equations
Learn systematic methods for solving trigonometric equations, finding general solutions, and determining all solutions within restricted domains.
The Solution Process
Unlike algebraic equations, trigonometric equations typically have infinitely many solutions because trig functions are periodic. To solve them systematically:
Isolate the trig function: Rearrange so you have sin θ = k, cos θ = k, or tan θ = k.
Find the reference angle: Determine the acute angle α where the trig ratio equals |k|.
Determine the quadrants: Use the ASTC rule to find which quadrants give the correct sign.
Write all solutions in the given domain (or the general solution using + 2nπ or + nπ).
Solution Patterns (General Solutions)
| Equation | General Solution |
|---|---|
| sin θ = k | θ = α + 2nπ or θ = (π − α) + 2nπ |
| cos θ = k | θ = α + 2nπ or θ = (2π − α) + 2nπ, i.e. θ = ±α + 2nπ |
| tan θ = k | θ = α + nπ |
where α is the reference angle and n is any integer
Working with Restricted Domains
Most HSC questions ask you to find solutions in a specific interval, such as 0 ≤ θ ≤ 2π or 0° ≤ θ ≤ 360°. After finding the general solutions, you substitute values of n to find all solutions that lie within the given domain.
Common Mistake to Avoid
Many students find only one solution when there are two (or more) in the given domain. Always check all quadrants and verify that you have found every solution in the domain.
Equations Requiring Factorisation
Some trig equations involve squared terms or products. These require factorisation before solving:
Type 1: Factor out common term
2 sin θ cos θ = 0
sin θ = 0 or cos θ = 0
Type 2: Quadratic in trig function
2 sin² θ − sin θ − 1 = 0
Let u = sin θ: 2u² − u − 1 = 0
Key Vocabulary
General Solution
A formula giving all possible solutions using an integer parameter n.
Restricted Domain
A specified interval (e.g. 0 ≤ x ≤ 2π) in which solutions must lie.
Reference Angle
The acute angle used to find solutions in all relevant quadrants.
Extraneous Solution
A solution that does not satisfy the original equation, often produced by squaring both sides.
Worked Examples
Solve sin θ = 1/2 for 0 ≤ θ ≤ 2π.
Step 1: Reference angle: α = π/6 (since sin(π/6) = 1/2).
Step 2: Sin is positive in Q1 and Q2.
Step 3: Q1: θ = π/6. Q2: θ = π − π/6 = 5π/6.
Answer: θ = π/6, 5π/6
Solve 2 cos θ + 1 = 0 for 0 ≤ θ ≤ 2π.
Step 1: Isolate: cos θ = −1/2.
Step 2: Reference angle: α = π/3 (since cos(π/3) = 1/2).
Step 3: Cos is negative in Q2 and Q3.
Answer: θ = 2π/3, 4π/3
Solve tan θ = −1 for 0 ≤ θ ≤ 2π.
Step 1: Reference angle: α = π/4 (since tan(π/4) = 1).
Step 2: Tan is negative in Q2 and Q4.
Step 3: Q2: θ = π − π/4 = 3π/4. Q4: θ = 2π − π/4 = 7π/4.
Answer: θ = 3π/4, 7π/4
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
How many solutions does sin θ = √3/2 have in the interval 0 ≤ θ ≤ 2π?
Question 2
Solve cos θ = 0 for 0 ≤ θ ≤ 2π.
Question 3
What is the general solution of tan θ = 1?
Question 4
Solve 2 sin θ − 1 = 0 for 0 ≤ θ ≤ 2π.
Question 5
How many solutions does sin θ = 2 have?
Key Concepts Summary
- ●Trig equations have multiple solutions because trig functions are periodic.
- ●Find the reference angle first, then use ASTC to determine quadrants.
- ●For tan, the general solution uses + nπ; for sin and cos, use + 2nπ.
- ●Always check the domain to list only the required solutions.
- ●If |k| > 1 for sin θ = k or cos θ = k, there are no solutions.