Solving trigonometric equations on stated intervals | Year 11

01 / LEARN

Understand the idea.

Use the full interval

A principal inverse-trigonometric answer is only a starting point. Use quadrant signs and symmetry to find all angles in one cycle, then repeat them through the stated interval. Check included endpoints separately so a boundary root is neither lost nor duplicated.

When the argument is a multiple of the variable, change its interval as well. For example, doubling $x$ doubles both endpoints. Solve for that full argument first, then undo the multiplication for every candidate.

Do algebra first

Simplify the equation algebraically before finding angles. A quadratic in sine or cosine can be factored by treating the ratio as one unknown. Reject ratio values outside $[- 1, 1]$ , and avoid dividing by a trigonometric factor unless its zero cases are checked.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Solve $sin x = \sqrt{3} / 2$ on $0 \leq x \leq 2 π$ .

Worked solution

1
Find the reference angle.
$sin \frac{π}{3} = \frac{\sqrt{3}}{2}$
The exact sine value identifies the acute reference angle.
2
Use the positive-sine quadrants.
$x = \frac{π}{3} or x = π - \frac{π}{3} = \frac{2 π}{3}$
Sine is positive in quadrants I and II.
3
Check the interval and endpoints.
$sin 0 = sin 2 π = 0$
Both candidates lie in $[0, 2 π]$ , and neither included endpoint supplies another root.

Answer

$x = \frac{π}{3}, \frac{2 π}{3}$

EXAMPLE 02

Worked example 2

Solve $cos (2 x) = 1 / 2$ on $0 \leq x \leq 2 π$ .

Worked solution

1
Change variables and the interval together.
$u = 2 x, 0 \leq u \leq 4 π$
The argument spans two full cycles as the original variable spans one.
2
Find the roots in the first cycle.
$u = \frac{π}{3}, \frac{5 π}{3}$
Cosine is one half in quadrants I and IV.
3
Include the second cycle.
$u = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3}, \frac{11 π}{3}$
Adding $2 π$ gives the further roots in $[0, 4 π]$ ; its endpoints have cosine 1.
4
Undo the change of variable.
$x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$
Dividing every argument value by 2 returns all roots to the requested variable and interval.

Answer

$x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve cosx=−1 on $0 \leq x \leq 2 π$ .

Show answer & reasoning

x=π.

Cosine equals $- 1$ at the leftmost unit-circle point. On $0 \leq x \leq 2 π$ , this happens only at $x = π$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Solve $sin x = \frac{1}{2}$ on $0 \leq x \leq 2 π$ .
3 marks
Solve $cos x = - \frac{\sqrt{2}}{2}$ on $0 \leq x \leq 2 π$ .
3 marks
Solve $tan x = 1$ on $- π \leq x \leq π$ .
3 marks
Solve $sin (2 x) = 0$ on $0 \leq x \leq 2 π$ .
3 marks
Solve $2 {sin}^{2} x - sin x = 0$ on $0 \leq x < 2 π$ .
3 marks
Solve $cos (x - \frac{π}{4}) = 0$ on $0 \leq x \leq 2 π$ .
3 marks

Read the worked solutions

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

Solve $sin x = \frac{1}{2}$ on $0 \leq x \leq 2 π$ .
3 marks
Use positive sine in quadrantsI andII.
$\frac{π}{6}, \frac{5 π}{6}$
Solve $cos x = - \frac{\sqrt{2}}{2}$ on $0 \leq x \leq 2 π$ .
3 marks
Use reference $\frac{π}{4}$ in quadrantsII andIII.
$\frac{3 π}{4}, \frac{5 π}{4}$
Solve $tan x = 1$ on $- π \leq x \leq π$ .
3 marks
Tangent repeats after $π$ .
$- \frac{3 π}{4}, \frac{π}{4}$
Solve $sin (2 x) = 0$ on $0 \leq x \leq 2 π$ .
3 marks
$2 x$ runs from 0 to $4 π$ ; its zeros are multiples of $π$ .
$0, \frac{π}{2}, π, \frac{3 π}{2}, 2 π$
Solve $2 {sin}^{2} x - sin x = 0$ on $0 \leq x < 2 π$ .
3 marks
$sin x (2 sin x - 1) = 0$ .
Combine the $sin x = 0$ and $sin x = \frac{1}{2}$ roots; exclude $2 π$ .
$0, \frac{π}{6}, \frac{5 π}{6}, π$
Solve $cos (x - \frac{π}{4}) = 0$ on $0 \leq x \leq 2 π$ .
3 marks
$x - \frac{π}{4} = \frac{π}{2} + k π$ ; retain the two allowed values.
$\frac{3 π}{4}, \frac{7 π}{4}$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: triangle rules, circular measure and trigonometric functions; school sequencing may vary.

Official curriculum source ↗

In this lesson

Understand the idea.

Use the full interval

Do algebra first

See the method in action.

Worked example 1

Find the reference angle.

Use the positive-sine quadrants.

Check the interval and endpoints.

Worked example 2

Change variables and the interval together.

Find the roots in the first cycle.

Include the second cycle.

Undo the change of variable.

Try it yourself.

Check your understanding

Keep practising