7.5Solving Trigonometric Equations
Solve linear trig equations (e.g. 2sinθ − 1 = 0) and quadratic trig equations. Find all solutions on [0, 2π) and write the general solution using periodicity.
Solving trig equations is a core algebraic skill that appears in physics (finding when a wave reaches a given value), engineering, and calculus (finding critical points of trig functions).
How do you find all angles that satisfy a trigonometric equation, given that trig functions are periodic and each value is achieved in multiple quadrants?
Solving a trig equation means finding all values of the variable that make the equation true. Because trig functions are periodic, there are infinitely many solutions — but they follow a predictable pattern.
General Solution Pattern
If sinθ = k, then:
θ = arcsin(k) + 2πn
θ = π − arcsin(k) + 2πn
If cosθ = k, then:
θ = ±arccos(k) + 2πn
If tanθ = k, then:
θ = arctan(k) + πn
Strategy
- Isolate the trig function.
- Find the reference angle.
- Determine which quadrants apply.
- List all solutions on [0, 2π).
- Add the period (2π or π) for the general solution.
Solve: 2sinθ − 1 = 0 on [0, 2π)
Isolate sinθ: sinθ = 1/2.
Reference angle: arcsin(1/2) = π/6.
sinθ is positive in QI and QII.
QI: θ = π/6. QII: θ = π − π/6 = 5π/6.
Solve: √3 tanθ + 1 = 0 on [0, 2π)
Isolate tanθ: tanθ = −1/√3.
Reference angle: arctan(1/√3) = π/6.
tanθ is negative in QII and QIV.
QII: θ = π − π/6 = 5π/6. QIV: θ = 2π − π/6 = 11π/6.
Solve: 2cos²θ − cosθ − 1 = 0 on [0, 2π)
Factor as a quadratic in cosθ: (2cosθ + 1)(cosθ − 1) = 0.
Case 1: cosθ = 1 → θ = 0.
Case 2: cosθ = −1/2 → reference angle = π/3.
cosθ is negative in QII and QIII: θ = 2π/3, 4π/3.
Solve: sin(2θ) = √3/2 on [0, 2π)
Let u = 2θ. Since θ ∈ [0, 2π), u ∈ [0, 4π).
sin(u) = √3/2 → reference angle = π/3.
Solutions for u in [0, 4π): u = π/3, 2π/3, π/3 + 2π = 7π/3, 2π/3 + 2π = 8π/3.
Divide by 2: θ = π/6, π/3, 7π/6, 4π/3.
Solve: 2sin²θ − sinθ − 1 = 0 on [0, 2π)
Factor: (2sinθ + 1)(sinθ − 1) = 0.
Case 1: sinθ = 1 → θ = π/2.
Case 2: sinθ = −1/2 → reference angle = π/6.
sinθ is negative in QIII and QIV: θ = π + π/6 = 7π/6, θ = 2π − π/6 = 11π/6.
Solve: 2cosθ + √2 = 0 on [0, 2π)
Hint: Isolate cosθ, find the reference angle, then identify the two quadrants where cosine has that sign.
Solve: tan²θ − 3 = 0 on [0, 2π)
Hint: Solve for tanθ (two values: ±√3), then find all four solutions.
Solve: 2sin²θ + sinθ = 0 on [0, 2π)
Hint: Factor out sinθ first, then solve each factor.
Solve: cos(2θ) = 1/2 on [0, 2π)
Hint: Let u = 2θ, expand the domain to [0, 4π), find all u solutions, then divide by 2.
Write the general solution for: sinθ = −√3/2
Hint: Find the two solutions on [0, 2π), then add 2πn to each.
Interactive Practice — 5 Questions
The solutions of sinθ = 1/2 on [0, 2π) are:
To solve sin(2θ) = k on [0, 2π), the domain for 2θ is:
The general solution for cosθ = 0 is:
To solve 2cos²θ − cosθ − 1 = 0, the first step is:
How many solutions does sinθ = −1/2 have on [0, 2π)?
Common Mistakes
Finding only one solution (the reference angle) and forgetting the second quadrant solution.
Every trig equation with |k| < 1 has two solutions per period. Always check both quadrants where the sign is correct.
When solving sin(2θ) = k on [0, 2π), only looking for solutions in [0, 2π) for the argument 2θ.
If θ ∈ [0, 2π), then 2θ ∈ [0, 4π). You must find all solutions of sin(u) = k in [0, 4π).
Dividing both sides by sinθ (or cosθ) when solving a factored equation, losing solutions where sinθ = 0.
Factor instead of dividing. Dividing by a trig function discards the solutions where that function equals zero.
Writing the general solution for tanθ = k as θ = arctan(k) + 2πn.
Tangent has period π, not 2π. The general solution is θ = arctan(k) + πn.
Math Tips
Always isolate the trig function first before finding the reference angle. Never try to take arcsin of a non-isolated expression.
Draw a quick unit circle sketch and mark the quadrants where the trig function has the required sign. This prevents missing solutions.
For equations with double angles (sin(2θ), cos(2θ)), double the domain before solving, then halve all solutions at the end.
Tangent has period π, so its general solution adds πn. Sine and cosine have period 2π, so they add 2πn.
After finding solutions, always verify by substituting back into the original equation — especially for quadratic trig equations where extraneous solutions can appear.