Trig Equations & General Solutions — Grade 11 Maths (CAPS)

Finding the reference angle, writing the general solution with + k·360° or + k·180°, and then listing the specific solutions that fall inside a given interval.

sin, cos and tan repeat, so an equation like sinθ = 0,5 has infinitely many answers. The general solution captures all of them in one line.

1The three general solutions

For a positive value a, with ref its reference angle:
sinθ = a ⇒ θ = ref + k·360° or θ = 180° − ref + k·360°
cosθ = a ⇒ θ = ±ref + k·360°
tanθ = a ⇒ θ = ref + k·180°

k ∈ ℤ — k is any integer, a different letter from the value a

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sin gets two branches, cos gets ±, tan gets 180°. Tan only needs 180° because it repeats twice as often as the other two.

2Step 1 — the reference angle

Ignore the sign and use the inverse of the SAME ratio as the equation: ref = sin⁻¹|a| for a sine equation, ref = cos⁻¹|a| for a cosine equation and ref = tan⁻¹|a| for a tangent equation. The reference angle is always positive and acute.

Worked ExampleDetermine the general solution of sinθ = 0,5

1
ref = sin⁻¹(0,5) = 30°.
2
θ = 30° + k·360° or θ = 180° − 30° + k·360°.
sin needs both branches.
3
θ = 30° + k·360° or θ = 150° + k·360°, k ∈ ℤ.
Never leave off 'k ∈ ℤ' — it carries a mark.

3When the value is negative

Worked ExampleDetermine the general solution of cosθ = −0,5

1
ref = cos⁻¹(0,5) = 60°.
Use the POSITIVE value for the reference angle.
2
cos is negative in QII and QIII.
CAST: cos is positive only in QI and QIV.
3
In QII: 180° − 60° = 120°.
4
θ = ±120° + k·360°, k ∈ ℤ.
The ± covers QIII automatically (−120° = 240°).

4Listing solutions in an interval

Worked ExampleSolve sinθ = 0,5 for θ ∈ [0° ; 360°]

1
General solution: θ = 30° + k·360° or 150° + k·360°.
2
k = 0: θ = 30° and θ = 150°. Both are in range.
3
k = 1: 390° and 510° — too big. k = −1: −330° and −210° — too small.
Test values of k until you fall outside the interval.
4
θ = 30° or θ = 150°.

5When the angle is a compound expression

Worked ExampleDetermine the general solution of tan(2x − 30°) = 1

1
ref = tan⁻¹(1) = 45°.
2
2x − 30° = 45° + k·180°.
Solve for the whole bracket first.
3
2x = 75° + k·180°.
4
x = 37,5° + k·90°, k ∈ ℤ.
Divide EVERYTHING by 2 — including the k term.

⚠️

The most common error in this whole topic: dividing the angle but forgetting to divide k·180°. Writing x = 37,5° + k·180° loses half the solutions.

6When one side is sin and the other is cos

The general solutions above need the same ratio on both sides. Use the co-functions to make them match: sinθ = cos(90° − θ) and cosθ = sin(90° − θ).

Worked ExampleDetermine the general solution of cos 2θ = sinθ

1
sinθ = cos(90° − θ), so cos 2θ = cos(90° − θ).
Now both sides are cosines.
2
2θ = ±(90° − θ) + k·360°.
cos gets ±.
3
With +: 2θ = 90° − θ + k·360°, so 3θ = 90° + k·360° and θ = 30° + k·120°.
Divide every term by 3 — the k term too.
4
With −: 2θ = −90° + θ + k·360°, so θ = −90° + k·360°.
5
θ = 30° + k·120° or θ = −90° + k·360°, k ∈ ℤ.
Check θ = 30°: cos 60° = 0,5 = sin 30°.

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If the equation is sin = cos, turn the cos into a sine instead and use the two sine branches.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Determine the general solution of sinθ = 0,5
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θ = 30° + k·360° or θ = 150° + k·360°, k ∈ ℤ
2
Determine the general solution of cosθ = −0,5
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ref = 60°, cos negative in QII and QIII →θ = ±120° + k·360°, k ∈ ℤ
3
Determine the general solution of tanθ = √3
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θ = 60° + k·180°, k ∈ ℤ
4
Solve sinθ = 0,5 for θ ∈ [0° ; 360°]
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θ = 30° or θ = 150°
5
Solve cosθ = 0,7660 for θ ∈ [0° ; 360°] (round to one decimal)
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ref = 40,0° → θ = 40,0° or θ = 320,0°
6
Determine the general solution of tan(2x − 30°) = 1
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2x − 30° = 45° + k·180° → x = 37,5° + k·90°, k ∈ ℤ
7
State the reference angle for sinθ = 0,8660.
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ref = sin⁻¹(0,8660) = 60° (to the nearest degree)
8
Determine the general solution of sin 2θ = cosθ.
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cosθ = sin(90° − θ), so sin 2θ = sin(90° − θ) (1)
2θ = 90° − θ + k·360° ⇒ 3θ = 90° + k·360° ⇒ θ = 30° + k·120° (1)
or 2θ = 180° − (90° − θ) + k·360° ⇒ θ = 90° + k·360° (1)
θ = 30° + k·120° or θ = 90° + k·360°, k ∈ ℤ (1)
9
Why must k ∈ ℤ be written down?
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It states that the pattern repeats for every integer k — without it the solution is incomplete and loses a mark.