Solving trig equations and describing all solutions | Year 11

01 / LEARN

The idea, step by step

Principal value is only a start

A calculator inverse sine gives one principal angle, not every solution. Use unit-circle signs and symmetry to find the second branch. Check both endpoints when the required interval is closed.

For an argument $2 x$ with $0 \leq x < 2 π$ , solve in $0 \leq 2 x < 4 π$ before dividing by 2. Simply using one cycle in the argument would omit solutions.

sin x = sin α ⟹ x = α + 2 k π or x = π - α + 2 k π, k \in ℤ

Factor and preserve restrictions

A product is zero if at least one factor is zero. Factor a trigonometric equation rather than dividing by a sine or cosine factor, because that division might lose solutions where the factor is zero.

For quadratic expressions, substitute $u = sin x$ or $u = cos x$ . Reject algebraic roots outside $[- 1, 1]$ , then solve each remaining trigonometric equation. Check the original denominators before accepting any result.

A common trap

Dividing $sin x cos x = sin x$ by $sin x$ without checking its zeros.

Instead: Factor $sin x (cos x - 1) = 0$ , then solve each factor.

02 / WORKED EXAMPLES

See the thinking

Example 1

Solve $2 sin x = 1$ on $[0, 2 π)$ .

Divide by 2: $sin x = 1 / 2$ .
Reference angle $π / 6$ ; sine is positive in quadrants I and II.
Both $π / 6$ and $5 π / 6$ fit the interval.

x = π / 6, 5 π / 6

Example 2

Solve $cos (2 x) = 0$ on $[0, 2 π)$ .

The argument $u = 2 x$ lies in $[0, 4 π)$ .
Cosine is zero at $u = π / 2, 3 π / 2, 5 π / 2, 7 π / 2$ .
Divide all four results by 2.

x = π / 4, 3 π / 4, 5 π / 4, 7 π / 4

Check your understanding

Give every real solution of $tan x = 1$ .

Show answer

$x = π / 4 + k π$ , $k \in ℤ$ .

Tangent repeats every $π$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $cos x = - 1 / 2$ on $[0, 2 π)$ .
2 marks
Solve $sin x = 0$ on $[- π, π]$ .
2 marks
Give all real solutions of $cos x = 1$ .
2 marks
Solve $2 {cos}^{2} x - cos x - 1 = 0$ on $[0, 2 π)$ .
2 marks
Solve $sin (2 x) = 0$ on $[0, π]$ .
2 marks
Solve $sin x (2 cos x - 1) = 0$ on $[0, 2 π)$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $cos x = - 1 / 2$ on $[0, 2 π)$ .
2 marks
Reference angle $π / 3$ ; use quadrants II and III.
$2 π / 3, 4 π / 3$ .
Solve $sin x = 0$ on $[- π, π]$ .
2 marks
All integer multiples of $π$ in the closed interval count.
$- π, 0, π$ .
Give all real solutions of $cos x = 1$ .
2 marks
Cosine is 1 at complete turns.
$x = 2 k π$ , $k \in ℤ$ .
Solve $2 {cos}^{2} x - cos x - 1 = 0$ on $[0, 2 π)$ .
2 marks
Factor $(2 cos x + 1) (cos x - 1) = 0$ ; solve cosine $- 1 / 2$ or 1.
$0, 2 π / 3, 4 π / 3$ .
Solve $sin (2 x) = 0$ on $[0, π]$ .
2 marks
$2 x = k π$ with $k = 0, 1, 2$ .
$0, π / 2, π$ .
Solve $sin x (2 cos x - 1) = 0$ on $[0, 2 π)$ .
2 marks
Sine zero gives 0 and $π$ ; cosine $1 / 2$ gives the other two.
$0, π / 3, π, 5 π / 3$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Further Trigonometry: periodic functions and models, trigonometric identities and reciprocal functions; the Library sequence follows the approved school-topic structure.

Official curriculum source ↗

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