🧮 Part I — Trig Equations & General Solutions
sin(x) = ½ vs sin⁻¹(½) = x — Are They the Same?
Understanding why equations have more solutions than inverse trig functions
Equation 1 — sin(x) = ½
On the unit circle, two angles satisfy sin(x)=½ for 0≤x≤2π:
x = π/6 and x = 5π/6
Since sine is periodic (period 2π), there are infinitely many solutions overall.
x = π/6 and x = 5π/6
Since sine is periodic (period 2π), there are infinitely many solutions overall.
Equation 2 — sin⁻¹(½) = x
Due to the domain restriction of arcsin, there is only one answer:
x = π/6 only
The range of arcsin is [−π/2, π/2], so only the Q1 angle is valid.
x = π/6 only
The range of arcsin is [−π/2, π/2], so only the Q1 angle is valid.
These are NOT equivalent
sin(x) = ½ has two solutions per period and infinitely many overall. sin⁻¹(½) = x gives only one answer due to the restricted range. When solving equations, never just use inverse trig and stop — you must find ALL solutions in the given domain.
General Solutions
Infinitely many solutions — add period × k to each base solution
| Equation | Solution 1 (general) | Solution 2 (general) | Notes |
|---|---|---|---|
| sin(x) = ½ | x = π/6 + 2πk | x = 5π/6 + 2πk | Two solutions per period (Q1 and Q2) |
| cos(x) = ½ | x = π/3 + 2πk | x = 5π/3 + 2πk | Two solutions per period (Q1 and Q4) |
| tan(x) = 1 | x = π/4 + πk | — | Period = π → one equation captures all solutions |
Why does tan only need one equation?
tan has period π (half of sin/cos). The two unit circle solutions for tan are exactly π apart, so adding πk to one solution already captures both. For sin and cos (period 2π), the two solutions per period are not 2π apart, so we need two separate equations.
4-Step Method for Solving Trig Equations
Isolate → unit circle → domain → write solutions
🔑 Solving Trig Equations — Step by Step
1
Isolate the trig function on one side of the equation. Get it in the form
sin(x) = value, cos(x) = value, or tan(x) = value.2
Find the angles on the unit circle that satisfy the equation. Remember to check ALL quadrants that give the correct sign.
3
Apply domain restrictions. If [0, 2π): list only those solutions. If no restriction: write general solutions with +2πk (sin/cos) or +πk (tan).
4
Write solutions clearly with correct notation.
📌 Example 1 — Find all solutions to cos(x) = −√2/2.
1
Already isolated. Reference angle: cos = √2/2 at π/4.2
cos is negative → Q2 and Q3. Q2: π − π/4 = 3π/4. Q3: π + π/4 = 5π/4.3
No domain restriction → general solutions.x = 3π/4 + 2πk and x = 5π/4 + 2πk, where k is an integer
📌 Example 2 — Find all solutions to sin(x) = −1.
1
sin = −1 only at 3π/2 (bottom of unit circle). Only one solution per period.2
Add 2πk since period = 2π.x = 3π/2 + 2πk, where k is an integer
📌 Example 3 — Find all solutions to tan(x) = 1/√3.
1
tan = 1/√3 at π/6 (Q1). tan is positive → also in Q3, but since period = π, one equation covers both.2
Add πk (tan's period).x = π/6 + πk, where k is an integer
📌 Example 4 — Find all values of θ where 6sin(θ) = 3√2.
1
Isolate: sin(θ) = 3√2/6 = √2/2.2
sin = √2/2 → Q1 (π/4) and Q2 (3π/4). Add 2πk for general solutions.θ = π/4 + 2πk and θ = 3π/4 + 2πk, where k is an integer
📌 Example 5 — f(θ) = 3 + 2cosθ and g(θ) = 2. Find x-coordinates of intersections for 0 ≤ θ < 2π.
1
Set f = g: 3 + 2cosθ = 2 → 2cosθ = −1 → cosθ = −1/2.2
cos = −1/2 → Q2: 2π/3 and Q3: 4π/3. Both are in [0, 2π).x = 2π/3 and x = 4π/3
Trig Notation & Quadratic-Type Equations
sin²x notation · factor like a polynomial · find all solutions in [0, 2π)
Squaring trig functions — notation
To square sin(x), we do NOT write sin(x²) — that only squares x.
Correct: write (sin x)² = sin²x. For example, (sin x)² = sin²x.
This is the standard AP exam notation. The exponent goes on the function name, not inside the argument.
📌 Example 6 — g(θ) = 3 − 4sin²θ. Find zeros on [0, 2π).
1
Set g = 0: 3 − 4sin²θ = 0 → sin²θ = 3/4 → sinθ = ±√3/2.2
sinθ = √3/2: θ = π/3 (Q1), θ = 2π/3 (Q2).3
sinθ = −√3/2: θ = 4π/3 (Q3), θ = 5π/3 (Q4).θ = π/3, 2π/3, 4π/3, 5π/3
📌 Example 7 — Find all θ in [0, 2π) where cos²θ + cosθ = 0.
1
Factor: cosθ(cosθ + 1) = 0.2
cosθ = 0: θ = π/2, 3π/2.3
cosθ = −1: θ = π.θ = π/2, π, 3π/2
📌 Example 8 — f(x) = 2sin²θ and g(x) = √3 sinθ. Find x-coordinates of intersections on [0, 2π).
1
Set f = g: 2sin²θ = √3 sinθ → 2sin²θ − √3 sinθ = 0 → sinθ(2sinθ − √3) = 0.2
sinθ = 0: θ = 0, π.3
2sinθ − √3 = 0 → sinθ = √3/2: θ = π/3 (Q1), 2π/3 (Q2).θ = 0, π/3, 2π/3, π
📊 Part II — Trig Inequalities & Sign Charts
Inequalities from Graphs — Example 1
Read intervals directly from the graph where f(x) is above or below the x-axis
Graph of f on [−2π, 2π] — shaded region shows where f(x) ≤ 0
📌 Example 1 — Graph of f defined on [−2π, 2π]. Find all x where f(x) ≤ 0.
1
From the graph: f(x) ≤ 0 when the curve is on or below the x-axis.2
Reading the graph: x = −2π (single point), −π ≤ x ≤ 0 (full interval), and π ≤ x ≤ 2π (full interval).x = −2π OR −π ≤ x ≤ 0 OR π ≤ x ≤ 2π
Inequalities Using the Unit Circle — Examples 2–4
Identify where sin/cos is above/below a value · use unit circle arc region
Strategy for unit circle inequalities
sin θ ≥ value: find the two boundary angles (where sin = value exactly), then the arc where the y-coordinate is above that value.
cos θ > value: find where cos = value, then the arc where the x-coordinate is to the right of that value.
Think of it as: which part of the circle has y-coordinate (or x-coordinate) satisfying the inequality?
📌 Example 2 — Find all θ in [0, 2π) where sinθ ≥ √2/2.
1
Find where sinθ = √2/2: θ = π/4 and θ = 3π/4.2
sinθ ≥ √2/2 means y-coordinate ≥ √2/2 on unit circle. This is the arc between π/4 and 3π/4 (top of circle).3
Include endpoints since ≥ (not strict >).π/4 ≤ θ ≤ 3π/4
📌 Example 3 — f(x)=4+2cosx, g(x)=3. Find all x in [0, 2π] where f(x) < g(x).
1
Set up: 4 + 2cosx < 3 → 2cosx < −1 → cosx < −1/2.2
Find where cosx = −1/2: x = 2π/3 and x = 4π/3.3
cos < −1/2 means x-coordinate < −1/2 → arc between 2π/3 and 4π/3 (left side of circle, strict inequality).2π/3 < x < 4π/3
📌 Example 4 — h(x)=3cosx, k(x)=√3+cosx. Find all x in [0, 2π] where h(x) > k(x).
1
Set up: 3cosx > √3 + cosx → 2cosx > √3 → cosx > √3/2.2
Find where cosx = √3/2: x = π/6 and x = 11π/6.3
cos > √3/2 means x-coordinate > √3/2 → arc near x=0 (right side). Includes x=0 and x=2π endpoints.0 ≤ x < π/6 OR 11π/6 < x ≤ 2π
Sign Charts with Trig Inequalities — Examples 5 & 6
Factor → find zeros → sign chart → read intervals
🔑 Solving Trig Inequalities with Sign Charts
1
Rearrange so one side = 0. Factor if needed.
2
Find zeros: set each factor = 0 and solve for θ in the given domain.
3
Build a sign chart: mark zeros on a number line. Test a value in each interval to find (+) or (−).
4
Signs alternate between zeros (unless a factor has even multiplicity). Label all intervals.
5
Read the answer: select intervals with the desired sign. Include endpoints if ≤ or ≥.
📌 Example 5 — Find all θ in [0, 2π] where 2sin²θ + sinθ < 0.
1
Factor: sinθ(2sinθ + 1) = 0.2
sinθ = 0: θ = 0, π, 2π.2sinθ + 1 = 0 → sinθ = −1/2: θ = 7π/6, 11π/6.
3
Zeros in order: 0, π, 7π/6, 11π/6, 2π. Build sign chart:Sign chart for sinθ(2sinθ + 1)
| Interval | (0, π) | (π, 7π/6) | (7π/6, 11π/6) | (11π/6, 2π) |
|---|---|---|---|---|
| sinθ sign | (+) | (−) | (−) | (+) |
| 2sinθ+1 sign | (+) | (−) | (+) | (+) |
| Product | + | + | − | + |
We need the product < 0 → the (−) intervals.
π < θ < 7π/6 OR 11π/6 < θ < 2π
📌 Example 6 — Find all θ in [0, 2π] where 2cos²θ + cosθ − 1 ≥ 0.
1
Factor: (2cosθ − 1)(cosθ + 1) = 0.2
2cosθ − 1 = 0 → cosθ = 1/2: θ = π/3, 5π/3.cosθ + 1 = 0 → cosθ = −1: θ = π.
3
Zeros in order: 0, π/3, π, 5π/3, 2π. Build sign chart:Sign chart for (2cosθ − 1)(cosθ + 1)
| Interval | [0, π/3] | (π/3, π) | (π, 5π/3) | [5π/3, 2π] |
|---|---|---|---|---|
| 2cosθ−1 sign | (+) | (−) | (−) | (+) |
| cosθ+1 sign | (+) | (+) | (−) | (+) |
| Product | + | − | + | + |
We need product ≥ 0 → (+) intervals AND the zeros (θ=π/3, π, 5π/3).
0 ≤ θ ≤ π/3 OR θ = π OR 5π/3 ≤ θ ≤ 2π
Quick Reference
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🔑 General Solution Rules
sin=v: two families + 2πk each
cos=v: two families + 2πk each
tan=v: ONE family + πk (period=π)
Quadratic trig: factor, then solve each factor
Inequalities: sign chart → pick correct intervals
⚠️ Common Mistakes
❌ sin(x)=½ has only one solution
sin(x)=½ has TWO solutions per period: π/6 and 5π/6. arcsin(½)=π/6 is just one — don't confuse equations with inverse trig.
❌ General solution for tan: two families + 2πk
tan has period π, so ONE family + πk captures all solutions. Don't use 2πk for tan.
❌ Dividing both sides by sinθ
If you divide sinθ(2sinθ−1)=0 by sinθ, you lose the solution sinθ=0. Always FACTOR — never divide by a trig function.
❌ Strict vs non-strict inequality endpoints
For < or >: open intervals (exclude boundary). For ≤ or ≥: closed intervals (include boundary = zero points).