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Learn · Mathematics · Grade 11 · Trigonometry

Trigonometric equations

Solving sin x = a, cos x = a, tan x = a on an interval and in general form, and equations that factor.

High school Lesson 28 of 31 Interactive figure Practice

First, the rules

One reference angle, the right quadrants

Solving sin x = ½ is two steps: find the acute angle whose sine is ½ (30°), then place it in every quadrant where sine is positive (I and II). Everything else is the same recipe.

★ Recipe
Examples 1, 2
1. Isolate the function. 2. Reference angle from the special-angle table (ignore the sign). 3. Quadrants from the sign: sin + in I, II; cos + in I, IV; tan + in I, III. 4. Write the angles in the asked interval.
★ Quadrant angles
Example 1
I: θ, II: 180° − θ, III: 180° + θ, IV: 360° − θ
No solution
sin x = 2, cos x = −1.5 — sine and cosine never leave [−1, 1]. Tangent takes every value.
General solution
Example 2
sin x = sin θ ⇒ x = θ + 360°k or x = 180° − θ + 360°k
cos x = cos θ ⇒ x = ±θ + 360°k
tan x = tan θ ⇒ x = θ + 180°k
k any integer.
★ Harder shapes
Example 3
sin 2x = ½: solve for 2x on a doubled interval, then halve. 2 sin²x − sin x = 0: factor as sin x (2 sin x − 1) = 0. cos²x = ¾: cos x = ±√3 / 2, both signs.

See it

Cross the curve with y = k

Choose sin, cos or tan and slide k. Every crossing on 0° ≤ x < 360° is a solution — two for sine and cosine (when |k| ≤ 1), two for tangent, none when the line is out of reach.

Step by step

Worked examples

1

Three basic equations

reference angle + quadrants

Solve on 0° ≤ x < 360°: (a) cos x = −√22 (b) 2 sin x − √3 = 0 (c) tan x = −1.

Ask yourself: Reference angle first (ignore the sign), then the sign tells you the quadrants.

Show the solution
  1. (a) cos 45° = √2 / 2; cosine negative in II and III: x = 135°, 225°.
  2. (b) sin x = √3 / 2; reference 60°, sine positive in I and II: x = 60°, 120°.
  3. (c) tan 45° = 1; tangent negative in II and IV: x = 135°, 315°.

Answer {135°, 225°} · {60°, 120°} · {135°, 315°}

2

General solution

all solutions

Give all solutions of sin x = −12 in degrees.

Ask yourself: Find the two angles in one turn, then add the period.

Show the solution
  1. Reference angle 30°; sine negative in III and IV: 210° and 330°.
  2. Period 360°: x = 210° + 360°k or x = 330° + 360°k, k ∈ ℤ.
  3. In radians: 7π/6 + 2πk or 11π/6 + 2πk.

Answer 210° + 360°k, 330° + 360°k

3

Factoring and a double angle

harder shapes

Solve on 0° ≤ x < 360°: (a) 2 cos²x − cos x = 0 (b) sin 2x = √32.

Ask yourself: (a) Factor out cos x — never divide by it, you would lose solutions. (b) Solve for 2x first on 0° ≤ 2x < 720°.

Show the solution
  1. (a) cos x (2 cos x − 1) = 0: cos x = 0 gives 90°, 270°; cos x = ½ gives 60°, 300°.
  2. (b) 2x has sine √3 / 2: 2x = 60°, 120°, 420°, 480° (two turns). Halve: x = 30°, 60°, 210°, 240°.

Answer (a) {60°, 90°, 270°, 300°} (b) {30°, 60°, 210°, 240°}

Practice: solve on one turn

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Basic equations on 0° ≤ x < 360° — pick the full solution set, and spot the ones with no solution. Your best score is kept in this browser only.

About this lesson

Trigonometric equations: solving sin x = a, cos x = a and tan x = a on 0° ≤ x < 360° with reference angles and quadrant signs, the general solution with + 360°k, equations that need a substitution or factoring, and why sin x = 2 has no solution. Worked examples and practice.

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