IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read

Linear Trigonometric Equations

A linear trig equation has the form sin x = k, cos x = k, or tan x = k — possibly with a transformed angle like (ax + b). The calculator gives you one solution; the unit circle (or trig graph) gives you the rest. Master the principal value plus the second-solution rule and any interval is fair game.

📘 What you need to know

Principal value: from sin⁻¹, cos⁻¹, or tan⁻¹ — exact if possible, decimal otherwise.
Second solution per period: sin → 180° − x; cos → 360° − x (or just −x); tan → x + 180°.
Periodicity: add or subtract 360° (sin/cos) or 180° (tan) to fill the interval.
For sin(ax + b) = k: substitute y = ax + b, transform the interval, solve for y, then back-substitute for x.
Interval transform direction: multiply by a first, then add b. The opposite order is the most common error.
Match calculator mode to the question — degrees → DEG, radians → RAD.
Conversions: 180° = π rad; 360° = 2π rad.

Basic equations: sin x = k, cos x = k, tan x = k

sin x = k

x and 180° − x

two per 360° (when −1 < k < 1)

cos x = k

x and 360° − x

two per 360°; or use ±x

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

For cos x = k, the second solution can be written as −x instead of 360° − x — same angle, just a quicker way to write it. Useful when the interval includes negatives.

Equations with sin(ax + b) = k

When the angle is transformed (e.g., sin(2x − 30°) = k), use a substitution to turn it into a basic equation.

Substitution Let y = ax + b ⇒ sin y = k (or cos, or tan)

The trickiest step is transforming the interval. If x is in [α, β], then y = ax + b is in [aα + b, aβ + b] — multiply by a first, then add b. After solving for y, back-substitute: x = (y − b) / a.

🧭 Recipe — solve sin/cos/tan(ax + b) = k

Substitute y = ax + b. The equation becomes sin y = k (or cos, or tan).
Transform the interval: multiply by a, then add b.
Find principal and second values for y, and add periods (360° for sin/cos, 180° for tan) until the new interval is fully covered.
Back-substitute: x = (y − b) / a for each value of y.
Check by substituting one or two answers into the original equation.

Worked examples

WE 1

Solve a basic sin equation

Solve the equation 3 sin x + 1 = 2 for 0° ≤ x ≤ 360°. Give answers to 3 s.f.

Step 1: Isolate sin x 3 sin x = 1 → sin x = 1/3 Step 2: Principal value x₁ = sin⁻¹(1/3) = 19.47° Step 3: Second solution — sin: 180° − x₁ x₂ = 180° − 19.47° = 160.53° Step 4: Both lie in [0°, 360°] x ≈ 19.5° or 161° (3 s.f.)

WE 2

Solve √2 cos x = −1 in radians

Solve the equation √2 cos x = −1 for 0 ≤ x ≤ 2π. Give answers in exact form.

Step 1: Isolate cos x cos x = −1/√2 = −√2/2 Step 2: Principal value (radian mode) x₁ = cos⁻¹(−√2/2) = 3π/4 Step 3: Second solution — cos: 2π − x₁ x₂ = 2π − 3π/4 = 5π/4 x = 3π4, 5π4 cos negative → solutions in Q2 and Q3 ✓

WE 3

Solve tan x = −√3 in a large interval

Solve the equation tan x = −√3 for 0° ≤ x ≤ 720°.

Step 1: Principal value x₁ = tan⁻¹(−√3) = −60° Step 2: Tan repeats every 180° → keep adding 180° to land in [0°, 720°] −60° + 180° = 120° 120° + 180° = 300° 300° + 180° = 480° 480° + 180° = 660° 660° + 180° = 840° → outside x = 120°, 300°, 480°, 660° interval length 720° = 4 periods → 4 solutions ✓

WE 4

Solve sin(2x + 30°) = √3/2

Solve the equation sin(2x + 30°) = √3/2 for 0° ≤ x ≤ 360°.

Step 1: Substitute y = 2x + 30° sin y = √3/2 Step 2: Transform interval — multiply by 2, then add 30° 0° ≤ x ≤ 360° → 30° ≤ y ≤ 750° Step 3: Solve sin y = √3/2 in [30°, 750°] principal: y = 60°; second: y = 180° − 60° = 120° add 360°: y = 420°, 480° add 720°: 780°, 840° — outside Step 4: Back-substitute x = (y − 30°)/2 y = 60° → x = 15° y = 120° → x = 45° y = 420° → x = 195° y = 480° → x = 225° x = 15°, 45°, 195°, 225° period of sin(2x + 30°) is 180° → 2 periods in 360° → 4 solutions ✓

WE 5

Solve 2 cos(x + π/6) = 1 in radians

Solve the equation 2 cos(x + π/6) = 1 for 0 ≤ x ≤ 2π. Give answers in exact form.

Step 1: Isolate and substitute y = x + π/6 cos y = 1/2 Step 2: Transform interval — add π/6 to each end 0 ≤ x ≤ 2π → π/6 ≤ y ≤ 13π/6 Step 3: Solve cos y = 1/2 principal: y = π/3; second: y = 2π − π/3 = 5π/3 add 2π → y = 7π/3 ≈ 2.33π → outside 13π/6 ≈ 2.17π Step 4: Back-substitute x = y − π/6 y = π/3 → x = π/3 − π/6 = π/6 y = 5π/3 → x = 5π/3 − π/6 = 9π/6 = 3π/2 x = π6, 3π2

WE 6

Solve tan(3x − π/4) = 1 in radians

Solve the equation tan(3x − π/4) = 1 for 0 ≤ x ≤ π. Give answers in exact form.

Step 1: Substitute y = 3x − π/4 tan y = 1 Step 2: Transform interval — multiply by 3, then subtract π/4 0 ≤ x ≤ π → −π/4 ≤ y ≤ 11π/4 Step 3: Solve tan y = 1 — period π principal: y = π/4 add π → y = 5π/4 add π → y = 9π/4 add π → y = 13π/4 → outside 11π/4 Step 4: Back-substitute x = (y + π/4)/3 y = π/4 → x = (π/2)/3 = π/6 y = 5π/4 → x = (3π/2)/3 = π/2 y = 9π/4 → x = (5π/2)/3 = 5π/6 x = π6, π2, 5π6 period of tan(3x − π/4) is π/3 → 3 periods in [0, π] → 3 solutions ✓

💡 Top tips

Always transform the interval correctly: multiply by a, then add b. The opposite order gives the wrong range every time.
Sketch the graph or unit circle with a horizontal line at y = k. Counts the solutions visually.
Estimate the number of solutions first: interval length ÷ period gives the number of full waves; multiply by 2 (sin/cos) or 1 (tan) for the count.
Match calculator mode. Degrees → DEG; radians → RAD. Mixing them is the silent killer.
Use exact values when k is a known fraction (1/2, √2/2, √3/2, etc.). Decimals lose marks.

⚠ Common mistakes

Stopping at the principal value. There’s always at least a second solution per period — never report just one without checking.
Wrong interval transform. For y = 2x + 30° on x ∈ [0°, 360°], y goes to [30°, 750°] — not [60°, 720°].
Wrong second-solution rule. Sin uses 180° − x; cos uses 360° − x (or −x); tan uses x + 180°. Mixing them is the common error.
Forgetting to back-substitute. After solving for y, divide by a and shift to recover x.
Including or excluding endpoints incorrectly. Read the inequality — is it ≤ or <?

Final note in this section: Quadratic Trigonometric Equations. Equations like 2 cos²x − cos x − 1 = 0 — substitute, factor as a quadratic, then solve each linear piece using the methods from this note.

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