AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/3/3-10)
Unit 3 · Topic 3.10
3.10 Trigonometric Equations and Inequalities
To solve a trig equation, isolate the trig function, find every solution in one period using the unit circle or an inverse trig function, then add multiples of the period. Inequalities work the same way, plus a check of which intervals satisfy them.
Key terms
- general solution
- solutions in one period
- inverse trig function
- trigonometric inequality
Solutions in one period
First isolate the trig function, just like solving for x. For 2 sin θ + 1 = 0, get sin θ = −1/2.
Then find every angle in one period, such as [0, 2π), that works. Sine and cosine usually give two angles per period, because the horizontal line y = k crosses each wave twice. Use reference angles and quadrant signs, or a calculator's inverse function plus symmetry.
- Sine: if α = arcsin k, the other solution in one period is π − α.
- Cosine: if α = arccos k, the other solution is 2π − α (or −α).
- Tangent: one solution per period of π, α = arctan k.
All solutions
Because trig functions are periodic, a trig equation usually has infinitely many solutions. Add any whole number of periods: for sine and cosine, write θ = α + 2πk; for tangent, θ = α + πk, where k is any integer.
If the inside has been transformed, like sin(2θ) = 1/2, solve for the inside first (2θ = π/6 + 2πk or 5π/6 + 2πk), then divide everything by 2. The solutions come π apart, not 2π, because the period of sin(2θ) is π.
Know how many solutions to expect. On [0, 2π), sin θ = k has two solutions when −1 < k < 1, one solution when k = 1 or k = −1, and none when k > 1 or k < −1. The same is true for cos θ = k.
Equations that need algebra first
Some equations are quadratic in a trig function. 2 cos² θ − cos θ − 1 = 0 factors like 2u² − u − 1 with u = cos θ: (2 cos θ + 1)(cos θ − 1) = 0. Then solve each factor.
Don't divide both sides by a trig function, which can throw away solutions where it equals 0. Move everything to one side and factor instead.
Inequalities and contexts
To solve an inequality like sin θ ≥ √3/2, solve the matching equation first to find the boundary angles. Then use the graph or unit circle to decide which intervals between them satisfy the inequality.
In a context, the domain is usually limited, such as 0 ≤ t ≤ 24 hours. That limits the number of solutions, so list only the ones inside the window.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
An equation and an inequality
(a) Solve 2 sin θ + 1 = 0 for 0 ≤ θ < 2π, and give all solutions. (b) Solve 2 sin θ + 1 < 0 for 0 ≤ θ < 2π.
Show the solutionHide the solution
- Step 1: (a) Isolate: sin θ = −1/2. The reference angle is π/6, and sine is negative in quadrants III and IV.
- Step 2: Quadrant III: π + π/6 = 7π/6. Quadrant IV: 2π − π/6 = 11π/6.
- Step 3: All solutions: θ = 7π/6 + 2πk or θ = 11π/6 + 2πk, for any integer k.
- Step 4: (b) sin θ < −1/2 where the sine graph dips below the line y = −1/2, which happens between the two boundary angles: 7π/6 < θ < 11π/6. (Test θ = 3π/2: sin = −1, which is less than −1/2.)
Answer: (a) θ = 7π/6, 11π/6 in [0, 2π); in general 7π/6 + 2πk and 11π/6 + 2πk. (b) 7π/6 < θ < 11π/6.
- Example 2
A quadratic in cosine
Solve 2 cos² θ − cos θ − 1 = 0 for 0 ≤ θ < 2π.
Show the solutionHide the solution
- Step 1: Factor: (2 cos θ + 1)(cos θ − 1) = 0.
- Step 2: 2 cos θ + 1 = 0 gives cos θ = −1/2, so θ = 2π/3 or 4π/3 (quadrants II and III, reference angle π/3).
- Step 3: cos θ − 1 = 0 gives cos θ = 1, so θ = 0.
- Step 4: Check one: at θ = 2π/3, 2(1/4) − (−1/2) − 1 = 1/2 + 1/2 − 1 = 0.
Answer: θ = 0, 2π/3, 4π/3.
- Example 3Calculator allowed
Trap: the calculator gives only one answer
Solve 3 cos x = 2 for 0 ≤ x < 2π.
Show the solutionHide the solution
- Step 1: cos x = 2/3. In radian mode, arccos(2/3) ≈ 0.841. That's the quadrant I solution, and it's the only one the calculator gives.
- Step 2: Cosine is also positive in quadrant IV. By symmetry, the other solution is 2π − 0.841 ≈ 5.442.
- Step 3: Stopping after the first value loses half the answer.
Answer: x ≈ 0.841 and x ≈ 5.442.
Common mistakes
- Stopping at the one angle an inverse trig function returns. Use symmetry to find the other solution in the period.
- Dividing by a trig function, such as sin θ, and losing the solutions where it equals 0.
- Adding 2πk to tangent solutions. Tangent's period is π, so add πk.
- Solving sin(2θ) = k and then forgetting to divide the added period by 2.
On the exam
- No-calculator questions expect exact angles from the unit circle; calculator questions expect three-decimal answers and every solution in the stated interval.
- In context questions, give every time in the window when the model hits the target value, and say what those times mean.
Connected topics
Videos
Check yourself
4 questions on 3.10 Trigonometric Equations and Inequalities. Pick an answer to see if you got it, and why.
What are all solutions to 2 cos θ + √3 = 0 for 0 ≤ θ < 2π?
For 0 ≤ θ < 2π, which of the following gives all values of θ for which sin θ > 1/2?
The depth of water at a dock, in meters, is modeled by D(t) = 2 cos(0.5t) + 6, where t is the number of hours after midnight. What is the first time t > 0 at which the depth is 6.8 meters?
Which of the following gives all solutions to tan θ = −1? (k is any integer.)
0 of 4 answered