AQA Level 2 Further Maths · Level 2 Certificate · Chapter 6: Geometry I · Lesson 6

Solving trigonometric equations

About 80 minutesCalculator allowed71 marks

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Date

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Four questions to start. Two of them use what you need today.

A1
Needed today
Solve $6 y^{2} + y - 2 = 0$ .
[2 marks]
Answer $y = \dots\dots$ or $y = \dots\dots$
A2
Last lesson
You are given that $cos 2 5^{\circ} = c$ . Write down two angles $x$ , with $0^{\circ} \leq x \leq 36 0^{\circ}$ , for which $cos x = - c$ .
[2 marks]
Answer $x = \dots\dots$ and $x = \dots\dots$
A3
From chapter 4
Solve $x^{2} - 3 x - 5 = 0$ . Give your answers to 2 decimal places.
[2 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$
A4
From chapter 6
Triangle $A B C$ has a right angle at $B$ . $A B = 7$ cm and $B C = 4$ cm. Work out angle $B A C$ . Give your answer to 1 decimal place.
[2 marks]
Answer

Example, then your turn

Level 2

Your teacher works through each example with you. Then try the one beside it. Give answers to 1 decimal place where they are not exact.

Example 1

Solve $cos x = - 0.35$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint The calculator gives the principal value. The cosine graph is symmetric about

x = 18 0^{\circ}

, so the second angle is

36 0^{\circ}

minus the first.

B1
Solve $sin x = - 0.45$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[2 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$

Example 2

Solve $4 tan x + 7 = 0$ for $- 36 0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint Make

tan x

the subject first. The tangent repeats every

18 0^{\circ}

: add and take away

18 0^{\circ}

until you leave the interval.

B2
Solve $9 sin x + 4 = 0$ for $0^{\circ} \leq x \leq 72 0^{\circ}$ .
[3 marks]
Answer $x = \dots\dots$

Example 3

Solve $5 tan^{2} x = 2$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint

tan^{2} x = 0.4

, so

tan x = 0.4

tan x = - 0.4

. Solve each one.

B3
Solve $9 sin^{2} x = 5$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[3 marks]
Answer $x = \dots\dots$

Example 4

Solve $2 cos^{2} x - 3 cos x - 2 = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint Write

c = cos x

and factorise

2 c^{2} - 3 c - 2

. A value of

cos x

outside

- 1

1

gives no solutions.

B4
Solve $6 sin^{2} x + sin x - 2 = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[4 marks]
Answer $x = \dots\dots$

Example 5

$f (t) = 2 t^{3} - 3 t^{2} - 3 t + 2$ . Show that $f (2) = 0$ . Hence solve $2 sin^{3} x - 3 sin^{2} x - 3 sin x + 2 = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint

(t - 2)

is a factor. Divide to find the quadratic factor, factorise it, then replace

t

sin x

B5
$g (t) = 3 t^{3} - 7 t^{2} - 7 t + 3$ . Show that $g (3) = 0$ . Hence solve $3 cos^{3} x - 7 cos^{2} x - 7 cos x + 3 = 0$ for $- 18 0^{\circ} \leq x \leq 18 0^{\circ}$ .
[5 marks]
Answer $x = \dots\dots$

Practice

Level 2

C1
Solve $(2 sin x - 1) (cos x + 1) = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[3 marks]
Answer $x = \dots\dots$
C2
Solve $2 tan^{2} x = 5 tan x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[4 marks]
Answer $x = \dots\dots$

C3
Do not use a calculator. Solve $2 cos x + 3 = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[2 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$
C4
Do not use a calculator. Work out the value of $x$ , where $0^{\circ} \leq x \leq 9 0^{\circ}$ , for which $4 sin^{2} x = 3$ .
[2 marks]
Answer $x = \dots\dots$
C5
Without solving them, write down how many solutions each equation has for $0^{\circ} \leq x \leq 108 0^{\circ}$ . Give a reason for each.
(i) $5 cos x = 2$
(ii) $3 sin x + 4 = 0$
[2 marks]
Answer(i) $\dots$ (ii) $\dots$
C6
Solve $7 cos x + 2 = 0$ for $- 36 0^{\circ} \leq x \leq 0^{\circ}$ .
[3 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$

Exam-style questions

AQA exam style

D1
Mia solves $3 sin^{2} x = sin x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ . She divides both sides by $sin x$ to get $sin x = \frac{1}{3}$ , and gives the answers $x = 19. 5^{\circ}$ and $x = 160. 5^{\circ}$ .
Explain why Mia's answer is not complete. Write down the other solutions.
[3 marks]
Answerother solutions $x = \dots\dots$

D2
One solution of $tan 2 x = 1$ is $x = 22. 5^{\circ}$ . Circle another solution.
$4 5^{\circ} 67. 5^{\circ} 112. 5^{\circ} 157. 5^{\circ}$
[1 mark]
Answer
D3
Solve $3 cos^{2} x + 5 cos x - 1 = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ . Give your answers to 1 decimal place. You must show your working.
[5 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$
D4
Answer both parts.
(a)
Factorise $4 y^{2} - 9 y + 2$ .
[2 marks]
Answer
(b)
Hence solve $4 tan^{2} x - 9 tan x + 2 = 0$ for $- 18 0^{\circ} \leq x \leq 18 0^{\circ}$ . Give your answers to 1 decimal place.
[4 marks]
Answer $x = \dots\dots$
Total for D4: 6 marks
D5
A seat on a fairground wheel is $h$ metres above the ground, where
$h = 12 - 10 cos x$
and $x^{\circ}$ is the angle the wheel has turned through since the seat was at the bottom.
(a)
Write down the greatest height of the seat.
[1 mark]
Answer $\dots\dots$ m
(b)
Work out the values of $x$ , for $0 \leq x \leq 360$ , for which the seat is $18$ m above the ground. Give your answers to 1 decimal place.
[3 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$
(c)
The wheel turns through two full turns, so $0 \leq x \leq 720$ . Work out the largest value of $x$ for which the seat is $18$ m above the ground.
[1 mark]
Answer $x = \dots\dots$
Total for D5: 5 marks
D6
Jo says, "The equation $4 sin x + 7 = 2$ has two solutions for $0^{\circ} \leq x \leq 36 0^{\circ}$ ."
Is Jo correct? Give a reason for your answer.
[2 marks]
Answer $\dots\dots$
D7
Do not use a calculator. Solve $2 sin^{2} x - sin x - 1 = 0$ for $36 0^{\circ} \leq x \leq 72 0^{\circ}$ . You must show your working.
[4 marks]
Answer $x = \dots\dots$

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
Solve $cos^{2} (2 x) = \frac{1}{4}$ for $0^{\circ} \leq x \leq 18 0^{\circ}$ . Do not use a calculator.
[4 marks]
Hint 1 · What to try
Let $u = 2 x$ . If $x$ goes from $0^{\circ}$ to $18 0^{\circ}$ , $u$ goes from $0^{\circ}$ to $36 0^{\circ}$ .
Hint 2 · The first line
$cos u = \frac{1}{2}$ or $cos u = - \frac{1}{2}$ .
Hint 3 · The full method
$u = 60$ , $300$ or $u = 120$ , $240$ . Halve each one: $x = 3 0^{\circ}$ , $6 0^{\circ}$ , $12 0^{\circ}$ , $15 0^{\circ}$ .
Answer $x = \dots\dots$

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can find all the solutions of sin x = k, cos x = k or tan x = k in an interval.
I can solve quadratic and cubic equations in sin x, cos x or tan x.
I can tell when an equation has no solutions, and not lose solutions by dividing.

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