A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 7: Trigonometry and modelling · Lesson 3

Solving trigonometric equations

115 marks

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Date

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Quick questions to start.

A1
Needed today
Expand $sin (x + 3 0^{\circ})$ , giving exact values for $sin 3 0^{\circ}$ and $cos 3 0^{\circ}$ .
[2 marks]
Answer
A2
From chapter 6
Solve $sec x = 2$ for $0^{\circ} \leq x < 36 0^{\circ}$ .
[2 marks]
Answer
A3
Common slip
Solve $tan x = - 1$ for $0^{\circ} \leq x < 36 0^{\circ}$ .
[2 marks]
Answer
A4
Today's idea
Solve $cos 2 x = \frac{1}{2}$ for $0^{\circ} \leq x < 18 0^{\circ}$ .
[3 marks]
Answer

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it.

Example 1

Solve $sin (x + 3 0^{\circ}) = 2 cos x$ for $0^{\circ} \leq x < 36 0^{\circ}$ .

Hint Expand the left side, put the sin x terms on one side and the cos x terms on the other, then divide by cos x.

B1
Solve $2 sin (x - 6 0^{\circ}) = cos x$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Give your answers to 1 decimal place.
[5 marks]
Answer

Example 2

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

Hint Use cos 2x = 2 cos²x − 1, so that only cos x is left.

B2
Solve $5 cos 2 x + 3 sin x = 1$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Give your answers to 1 decimal place where needed.
[5 marks]
Answer

Example 3

Solve $2 tan^{2} x + sec x = 1$ for $0^{\circ} \leq x < 36 0^{\circ}$ .

Hint Use tan²x = sec²x − 1, so that only sec x is left.

B3
Solve $cosec^{2} θ - 3 cot θ - 5 = 0$ for $0^{\circ} \leq θ < 36 0^{\circ}$ . Give your answers to 1 decimal place where needed.
[5 marks]
Answer

Example 4

Solve $sin (2 x + 3 0^{\circ}) = \frac{1}{2}$ for $0^{\circ} \leq x < 36 0^{\circ}$ .

Hint Find the interval for 2x + 30° first. It runs from 30° to 750°.

B4
Solve $cos (2 x - 4 0^{\circ}) = 0.3$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Give your answers to 1 decimal place.
[5 marks]
Answer

Example 5

Solve $4 sin x cos x = 3$ for $0 \leq x < 2 π$ .

Hint 4 sin x cos x is 2 sin 2x. Then 2x runs from 0 to 4π.

B5
Solve $cos^{2} x - sin^{2} x = sin x cos x$ for $0 \leq x < 2 π$ . Give your answers to 3 significant figures.
[5 marks]
Answer

Practice

A-Level

C1
Mia solves $sin 2 x = 0.6$ for $0^{\circ} \leq x < 36 0^{\circ}$ . She writes $2 x = 36. 9^{\circ}$ or $143. 1^{\circ}$ , so $x = 18. 4^{\circ}$ or $71. 6^{\circ}$ . Explain what she has missed, and give all the solutions to 1 decimal place.
[4 marks]
Answer

C2
Solve $3 sec^{2} x - 5 tan x - 5 = 0$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Give your answers to 1 decimal place.
[5 marks]
Answer
C3
Solve $cos (x + 6 0^{\circ}) = sin x$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Show that $tan x = 2 - 3$ and hence give exact answers.
[4 marks]
Answer
C4
Solve $cot x = 2 cos x$ for $0^{\circ} < x < 36 0^{\circ}$ .
[4 marks]
Answer
C5
Solve $cosec 2 x = 4$ for $0 \leq x < π$ . Give your answers to 3 significant figures.
[4 marks]
Answer

Exam-style questions

Edexcel exam style

D1
(a)
Show that $2 sin (x + 3 0^{\circ}) = 3 sin x + cos x$ .
[2 marks]
(b)
Hence solve $2 sin (x + 3 0^{\circ}) = 3 cos x$ for $0^{\circ} \leq x < 36 0^{\circ}$ , giving your answers to 1 decimal place.
[4 marks]
Answer
(c)
Hence solve $2 sin (2 θ + 3 0^{\circ}) = 3 cos 2 θ$ for $0^{\circ} \leq θ < 18 0^{\circ}$ , giving your answers to 1 decimal place.
[3 marks]
Answer
Total for D1: 9 marks

D2
Sam writes: " $sec^{2} x = tan^{2} x - 1$ , so $tan^{2} x - 2 tan x - 5 = 0$ and $tan x = 1 \pm 6$ ."
Sam is asked to solve $sec^{2} x = 2 tan x + 4$ for $0^{\circ} \leq x < 36 0^{\circ}$ .
(a)
Explain the mistake in Sam's working.
[2 marks]
(b)
Solve the equation correctly, giving your answers to 1 decimal place where needed.
[6 marks]
Answer
Total for D2: 8 marks
D3
Solve $cos (x + 4 5^{\circ}) + cos (x - 4 5^{\circ}) = sin 2 x$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Give exact answers.
[7 marks]
Answer
D4
(a)
Show that $\frac{1}{1 - sin x} + \frac{1}{1 + sin x} = 2 sec^{2} x$ , for $sin x \neq = \pm 1$ .
[3 marks]
(b)
Hence solve $\frac{1}{1 - sin x} + \frac{1}{1 + sin x} = 8$ for $0 \leq x < 2 π$ , giving exact answers.
[4 marks]
Answer
(c)
Explain why $\frac{1}{1 - sin x} + \frac{1}{1 + sin x} = 1$ has no solutions.
[2 marks]
Total for D4: 9 marks
D5
(a)
Show that the equation $tan (x + 4 5^{\circ}) = k tan x$ , where $k$ is a constant, can be written as $k tan^{2} x + (1 - k) tan x + 1 = 0$ .
[3 marks]
(b)
Solve $tan (x + 4 5^{\circ}) = 6 tan x$ for $0^{\circ} \leq x < 36 0^{\circ}$ , giving your answers to 1 decimal place.
[4 marks]
Answer
(c)
Find the set of values of $k$ for which $tan (x + 4 5^{\circ}) = k tan x$ has solutions. Give your answer in exact form.
[3 marks]
Answer
Total for D5: 10 marks
D6
(a)
Show that $sin (x + 6 0^{\circ}) + sin (x - 6 0^{\circ}) = sin x$ .
[2 marks]
(b)
Hence solve $sin (x + 6 0^{\circ}) + sin (x - 6 0^{\circ}) = 2 cos^{2} x$ for $0^{\circ} \leq x < 36 0^{\circ}$ , giving your answers to 1 decimal place.
[4 marks]
Answer
(c)
Hence solve $sin (2 θ + 6 0^{\circ}) + sin (2 θ - 6 0^{\circ}) = 2 cos^{2} 2 θ$ for $0^{\circ} \leq θ < 36 0^{\circ}$ , giving your answers to 1 decimal place.
[3 marks]
Answer
Total for D6: 9 marks

Extension

Stretch

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

E1
Solve $sin x + sin 2 x + sin 3 x = 0$ for $0 \leq x < 2 π$ . Give exact answers.
[8 marks]
Hint 1 · What to try
Leave $sin 2 x$ alone. Write $sin 3 x$ as $sin (2 x + x)$ and $sin x$ as $sin (2 x - x)$ .
Hint 2 · The first line
Add the two expansions: the $cos 2 x sin x$ terms cancel, leaving $2 sin 2 x cos x$ .
Hint 3 · The full method
Now $sin 2 x$ is a common factor. Do not divide by it: it can be $0$ .

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can expand a compound angle and collect the terms into tan x = k.
I can replace cos 2x or sec²x so that the equation has one function, and solve the quadratic.
I can solve an equation in 2x or (2x + 30°) by stretching the interval first.
I can factorise instead of dividing, so that no solution is lost.

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