Key Idea: Trig equations ask you to find every angle in a given domain that satisfies the equation — and because sin, cos and tan repeat, there is almost always more than one answer. Worth full marks on both papers.
🎯 The method: base angle, then all its partners
| Equation | First angle | Other solutions (per 360°) |
|---|---|---|
| sin x = k | x = sin⁻¹k | 180° − x |
| cos x = k | x = cos⁻¹k | 360° − x |
| tan x = k | x = tan⁻¹k | x + 180° |
Over one full turn, sin x = k and cos x = k (with |k| < 1) each give two solutions; tan x = k gives one per 180°. If k is negative, the angles sit in different quadrants — sketch the unit circle (or use CAST) to place them, e.g. cos x = −0.5 lands in Q2 and Q3.
🔁 Two cases that hide extra answers
| Type | What to do | Watch out |
|---|---|---|
| Multiple angle, e.g. sin(2x) = k | Solve over the stretched interval (0° ≤ 2x ≤ 720°), find all of those, then divide each by 2. | A multiple angle means more solutions — don't halve before you've found them all. |
| Quadratic in sin/cos, e.g. 2sin²x − sin x − 1 = 0 | Let s = sin x, factor/solve for s, then solve sin x = each value. | Reject any s outside [−1, 1]; if sin² and cos are mixed, swap with cos²x = 1 − sin²x first. |
✏️ IB-style worked examples
IB-style question — find all solutions in the domain (Paper 1)
Solve cos x = 0.5 for 0° ≤ x ≤ 360°, without a calculator.
Step by step:
Take the inverse for the base angle.
Second solution for cosine: 360° − x.
Both lie in 0° ≤ x ≤ 360°, so keep both.
x = 60° or x = 300°.
IB-style question — a multiple-angle equation (Paper 1)
Solve sin(2x) = 0.5 for 0° ≤ x ≤ 360°.
Step by step:
Let u = 2x, so the interval stretches to 0° ≤ u ≤ 720°.
All u in the doubled interval (base 30°, then 180°−30°, then +360°).
Divide each by 2 to get x.
x = 15°, 75°, 195°, 255° (four solutions).
IB-style question — a quadratic in sin x (Paper 1)
Solve 2sin²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°.
Step by step:
Let s = sin x and factor the quadratic.
So sin x = 1 or sin x = −½ (both are in [−1, 1]).
Solve each over the interval.
x = 90°, 210°, 330°.
IB-style question — solve inside a model with the GDC (Paper 2)
A tide height is h = 4 sin(30t)° + 6 (t in hours). Find the first time t > 0 when the height is 8 m.
Step by step:
Set up the equation.
Graph y = h and y = 8 on the GDC and use intersect over the interval.
Solve for the first time.
First at t = 1 hour (the GDC intersect confirms it).
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: The inverse on your calculator gives only one angle. The domain almost always wants more — find the partner (sin → 180°−x, cos → 360°−x, tan → +180°) and check it's inside the interval. For a multiple angle, find every solution in the stretched interval before you divide.
Tap each card to reveal the answer.
Solve sin x = 0.5 for 0° ≤ x ≤ 360° x = 30° or 150° — base angle 30°, partner 180° − 30°.
What is the second solution to cos x = k (in 0°–360°)? 360° − x — cosine is symmetric about 360°.
How many solutions does sin(2x) = k have on 0° ≤ x ≤ 360°? Up to four — solve over 0° ≤ 2x ≤ 720°, then divide by 2.
First step for 2cos²x − cos x = 0? Let c = cos x and factor: c(2c − 1) = 0, so cos x = 0 or ½.
cos x = −0.5 — which quadrants? Q2 and Q3 — x = 120° and 240°.
Exam tips
- Underline the domain first — it decides how many answers you keep.
- Inverse gives one angle; always find the partner (sin → 180°−x, cos → 360°−x, tan → +180°).
- Multiple angle: solve over the stretched interval, then divide — don't halve too early.
- Quadratic in sin/cos: substitute s = the ratio, factor, and reject any s outside [−1, 1].
- Paper 2: graph both sides and use intersect — perfect for 'when is the height …?' models.