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

Quadratic Trigonometric Equations

A quadratic trig equation has a squared trig term, like 2 cos²x + cos x − 1 = 0. Substitute y = cos x (or sin x, or tan x) and it becomes a normal quadratic. Solve for y, reject anything outside [−1, 1] for sin/cos, then solve each linear trig piece.

📘 What you need to know

Quadratics in a single ratio

Substitution Let y = sin x (or cos x, or tan x) → solve a y² + b y + c = 0

If the equation already contains only one trig function (squared and linear), the substitution is direct. Solve the quadratic by factoring or formula, reject any value of y that y can’t take, then solve each remaining linear trig equation.

Mixed ratios — convert first

Equation containsUseResult
sin²x and cos x (linear)sin²x = 1 − cos²xquadratic in cos x
cos²x and sin x (linear)cos²x = 1 − sin²xquadratic in sin x
tan x and sin/costan x = sin x / cos xmultiply through by cos x → factor
cos 2x and sin x or cos xcos 2x = 1 − 2 sin²x   or   2 cos²x − 1quadratic in sin x or cos x
Always swap the squared term to match the linear one. If the equation is “cos²x + linear sin x”, change the cos²x. The whole equation should end up in one trig function.

🧭 Recipe — solve a quadratic trig equation

  1. Reduce to one ratio: convert squares using sin²x + cos²x = 1, or replace tan x with sin x / cos x.
  2. Move everything to one side = 0.
  3. Substitute y = sin x (or cos x, or tan x) and solve the quadratic.
  4. Reject impossible values: |sin x| or |cos x| > 1.
  5. Solve each remaining linear equation in the given interval (principal + second + periods).

Worked examples

WE 1

Solve a quadratic in cos x

Solve the equation 2 cos²x + cos x − 1 = 0 for 0° ≤ x ≤ 360°.

Step 1: Substitute y = cos x 2y² + y − 1 = 0 Step 2: Factor (2y − 1)(y + 1) = 0 y = 1/2   or   y = −1 Step 3: Both values valid (in [−1, 1]) cos x = 1/2 → x = 60°, 300° cos x = −1 → x = 180° x = 60°, 180°, 300° cos x = −1 has only one solution per 360° (tangent point)
WE 2

Convert cos² to sin² then solve

Solve the equation 2 cos²x + 5 sin x = 4 for 0° ≤ x ≤ 360°.

Linear sin x → swap cos² for 1 − sin²x Step 1: Substitute 2(1 − sin²x) + 5 sin x = 4 2 − 2 sin²x + 5 sin x = 4 Step 2: Move to one side 2 sin²x − 5 sin x + 2 = 0 Step 3: Quadratic formula → sin x = (5 ± √(25 − 16))/4 = (5 ± 3)/4 sin x = 2 (reject, |sin x| ≤ 1) sin x = 1/2 Step 4: Solve sin x = 1/2 in [0°, 360°] x = 30°   or   x = 180° − 30° = 150° x = 30°, 150°
WE 3

Equation with tan x and sin x

Solve the equation 3 sin x = tan x for 0° ≤ x ≤ 360°. Give answers to 3 s.f. where necessary.

Step 1: Replace tan x with sin x / cos x 3 sin x = sin x / cos x Step 2: Multiply both sides by cos x and rearrange 3 sin x cos x = sin x 3 sin x cos x − sin x = 0 Step 3: Factor (do NOT divide by sin x) sin x (3 cos x − 1) = 0 sin x = 0   or   cos x = 1/3 Step 4: Solve each in [0°, 360°] sin x = 0 → x = 0°, 180°, 360° cos x = 1/3 → x = cos⁻¹(1/3) ≈ 70.53° or 360° − 70.53° = 289.47° x = 0°, 70.5°, 180°, 289°, 360° (3 s.f.) dividing by sin x would have killed the sin x = 0 solutions
WE 4

Solve a quadratic in sin x in radians

Solve the equation 2 sin²x − 3 sin x + 1 = 0 for 0 ≤ x ≤ 2π. Give answers in exact form.

Step 1: Substitute y = sin x 2y² − 3y + 1 = 0 Step 2: Factor (2y − 1)(y − 1) = 0 y = 1/2   or   y = 1 Step 3: Solve each in [0, 2π] sin x = 1/2 → x = π/6 or x = π − π/6 = 5π/6 sin x = 1 → x = π/2 x = π6, π2, 6 sin x = 1 has one solution per 2π (tangent point at top of unit circle)
WE 5

Solve a quadratic in tan x

Solve the equation tan²x − 4 tan x + 3 = 0 for 0° ≤ x ≤ 360°. Give answers to 3 s.f. where necessary.

Step 1: Substitute y = tan x y² − 4y + 3 = 0 Step 2: Factor (y − 1)(y − 3) = 0 y = 1   or   y = 3 Step 3: No rejection — tan x can take any value Step 4: Solve each — tan repeats every 180° tan x = 1 → x = 45°, 225° tan x = 3 → x = tan⁻¹(3) ≈ 71.57°, 251.57° x = 45°, 71.6°, 225°, 252° (3 s.f.)
WE 6

Quadratic formula when factoring fails

Solve the equation 2 sin²x + 3 sin x − 1 = 0 for 0° ≤ x ≤ 360°. Give answers to 3 s.f.

Step 1: Substitute y = sin x 2y² + 3y − 1 = 0 Step 2: Doesn’t factor — use the quadratic formula y = (−3 ± √(9 + 8)) / 4 = (−3 ± √17) / 4 y ≈ 0.281   or   y ≈ −1.781 Step 3: Reject y ≈ −1.781 (|sin x| ≤ 1) Step 4: Solve sin x ≈ 0.281 x = sin⁻¹(0.281) ≈ 16.31° second: 180° − 16.31° = 163.69° x ≈ 16.3°, 164° (3 s.f.) always check both roots before solving — a rejected root saves wasted work

💡 Top tips

⚠ Common mistakes

That closes Trigonometric Equations & Identities. You’ve now built every algebraic tool: identities, compound and double angle formulae, ratio relationships, and both linear and quadratic equations. The next section moves to Inverse & Reciprocal Trigonometric Functions — sec, cosec, cot, and their inverses, with their own graphs, identities, and equations.

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