IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read

Solving Equations Using Trigonometric Graphs

Your calculator gives you one solution to sin x = k — the principal value. The graph (or unit circle) gives you the rest. Sketch the curve, draw a horizontal line at y = k, and read off every intersection inside the required interval.

📘 What you need to know

Principal value: the answer the calculator gives via sin⁻¹, cos⁻¹, or tan⁻¹.
Second solution per period: sin → 180° − x₁; cos → 360° − x₁; tan → 180° + x₁.
Periodicity: add or subtract 360° (sin/cos) or 180° (tan) until you’ve covered the whole interval.
Number of solutions: divide the interval length by the period to get full periods, then multiply (sin/cos = 2 per period; tan = 1 per period).
Edge cases: k = 0 gives extra solutions; k = ±1 gives one solution per period (tangent point).
Match the units: degrees or radians — set your calculator before starting.

Finding all solutions

sin x = k

x and 180° − x

two per 360° (when −1 < k < 1)

cos x = k

x and 360° − x

two per 360° (when −1 < k < 1)

tan x = k x and x + 180° (one per 180°)

In radians: replace 360° with 2π and 180° with π. Same logic, different unit.

Counting solutions in an interval

Equation	Solutions per period	Period
sin x = k (−1 < k < 1, k ≠ 0)	2	360°
cos x = k (−1 < k < 1, k ≠ 0)	2	360°
tan x = k	1	180°
sin x = ±1 or cos x = ±1	1	360°
sin x = 0 or cos x = 0	2	360°

Quick estimate: divide the interval by the period. A non-whole answer means you’ll get the floor or ceiling number — sketch the graph to check exactly.

🧭 Recipe — solve sin/cos/tan = k in an interval

Principal value from the calculator.
Second solution in [0°, 360°]: sin → 180° − x₁; cos → 360° − x₁; tan → 180° + x₁.
Add and subtract 360° (or 180° for tan) to push the two base values into the rest of the interval.
Discard any solutions outside the interval.
List all answers in order, rounded as required.

Worked examples

WE 1

Solve sin x = 0.7 in degrees

Find all solutions of sin x = 0.7 in 0° ≤ x ≤ 360°. Give answers correct to 3 s.f.

Step 1: Principal value x₁ = sin⁻¹(0.7) = 44.43° Step 2: Second solution — sin: 180° − x₁ x₂ = 180° − 44.43° = 135.57° Step 3: Both lie in [0°, 360°] x ≈ 44.4° or 136° (3 s.f.)

WE 2

Solve cos x = −0.4

Find all solutions of cos x = −0.4 in 0° ≤ x ≤ 360°. Give answers correct to 3 s.f.

Step 1: Principal value x₁ = cos⁻¹(−0.4) = 113.58° Step 2: Second solution — cos: 360° − x₁ x₂ = 360° − 113.58° = 246.42° x ≈ 114° or 246° (3 s.f.) cos is negative in Q2 and Q3 — both solutions lie above 90° ✓

WE 3

Solve tan x = 2

Find all solutions of tan x = 2 in 0° ≤ x ≤ 360°. Give answers correct to 3 s.f.

Step 1: Principal value x₁ = tan⁻¹(2) = 63.43° Step 2: Second solution — tan: x₁ + 180° x₂ = 63.43° + 180° = 243.43° x ≈ 63.4° or 243° (3 s.f.) tan repeats every 180°, so only one solution per period

WE 4

Solve cos x = 0.3 in radians, larger interval

Find all solutions of cos x = 0.3 in 0 ≤ x ≤ 4π. Give answers correct to 3 s.f.

Step 1: Principal value (calculator in radian mode) x₁ = cos⁻¹(0.3) = 1.2661… Step 2: Second solution — cos: 2π − x₁ x₂ = 2π − 1.2661 = 5.0171… Step 3: Add 2π for the second period (4π ≈ 12.566) x₁ + 2π = 1.2661 + 6.2832 = 7.5493… x₂ + 2π = 5.0171 + 6.2832 = 11.300… Step 4: All four lie in [0, 4π] x ≈ 1.27, 5.02, 7.55, 11.3 (3 s.f.) interval of length 4π = two full periods → 4 solutions ✓

WE 5

Count solutions without solving

Without solving, find the number of solutions of cos x = 0.8 in the interval 0° ≤ x ≤ 1440°.

Step 1: Number of full periods 1440° ÷ 360° = 4 periods Step 2: 0 < 0.8 < 1 → 2 solutions per period total = 4 × 2 = 8 8 solutions draw the line y = 0.8 across four full waves to confirm

WE 6

Given one solution, find all others

One solution of sin x = 0.3 is x = 17.46°. Find all other solutions in the interval −180° ≤ x ≤ 540°. Give answers correct to 3 s.f.

Step 1: Second solution in [0°, 360°] — sin: 180° − x₁ 180° − 17.46° = 162.54° Step 2: Add 360° to reach the next period 17.46° + 360° = 377.46° 162.54° + 360° = 522.54° Step 3: Subtract 360° for the negative half of the interval 17.46° − 360° = −342.54° → outside [−180°, 540°] 162.54° − 360° = −197.46° → outside Step 4: Collect all solutions in interval, exclude given x = 17.46° x ≈ 163°, 377°, 523° (3 s.f.) total of 4 solutions in interval; excluding the given one leaves 3 others

💡 Top tips

Always sketch the graph with a horizontal line at y = k. Counts the solutions visually.
Match calculator mode to interval units. Degrees → DEG; radians → RAD.
Get both base solutions first in [0°, 360°], then shift by ±360° to fill the interval.
For tan, only one solution per 180° — half as many as sin/cos in any given interval.
Edge values matter: cos x = 1 has only one solution per 360° (at multiples of 360°). Same for sin x = 1 (at 90° + 360°k).

⚠ Common mistakes

Stopping at the principal value. Always give all solutions in the interval.
Wrong formula for the second solution. Sin uses 180° − x; cos uses 360° − x. Mixing them up gives wrong angles.
Forgetting negative parts of the interval. Subtract 360° (or 2π) to extend left.
Including endpoints incorrectly. Check whether the interval uses ≤ or <.
Using degrees on a radian calculator. sin(0.5) and sin(0.5°) are very different.

Next note: Transformations of Trigonometric Functions. Adding constants — like y = 2 sin(3x) − 1 — stretches and shifts the graph. Same shapes, new amplitudes, periods, and principal axes.

Need help with Solving Trig Equations?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →