A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 5: Radians · Lesson 3

Solving trigonometric equations in radians

112 marks

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Quick questions to start.

A1
Needed today
Solve $sin x = \frac{1}{2}$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[2 marks]
Answer
A2
Needed today
Write $22 5^{\circ}$ in radians, as a multiple of $π$ .
[2 marks]
Answer
A3
Needed today
Write $2 cos^{2} x + 3 sin x$ in terms of $sin x$ only.
[2 marks]
Answer
A4
Common slip
Given that $0 \leq x \leq 2 π$ , find the interval that $2 x + \frac{π}{3}$ lies in.
[2 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 $sin x = - 0.4$ for $0 \leq x \leq 2 π$ , giving your answers to 3 significant figures.

Hint The calculator gives one value, which may be outside the interval. Sine is negative in the third and fourth quarters, so use

π + α

and

2 π - α

where

α = sin^{- 1} 0.4

B1
Solve $cos x = - 0.3$ for $0 \leq x \leq 2 π$ , giving your answers to 3 significant figures.
[4 marks]
Answer

Example 2

Solve $tan (x + \frac{π}{6}) = 3$ for $0 \leq x \leq 2 π$ .

Hint Let

u = x + \frac{π}{6}

. Change the interval first:

\frac{π}{6} \leq u \leq \frac{13 π}{6}

. Solve for

u

, then take

\frac{π}{6}

away.

B2
Solve $2 cos (x - \frac{π}{4}) = - 1$ for $0 \leq x \leq 2 π$ , giving exact answers.
[4 marks]
Answer

Example 3

Solve $sin 2 x = \frac{3}{2}$ for $0 \leq x \leq π$ .

Hint If

0 \leq x \leq π

then

0 \leq 2 x \leq 2 π

. Find every value of

2 x

in that interval before you halve.

B3
Solve $cos 3 x = \frac{2}{2}$ for $0 \leq x \leq π$ , giving exact answers.
[5 marks]
Answer

Example 4

Solve $2 cos^{2} x + 3 sin x = 3$ for $0 \leq x \leq 2 π$ .

Hint Replace

cos^{2} x

1 - sin^{2} x

so that the equation has only

sin x

in it. Then it is a quadratic in

sin x

B4
Solve $2 sin^{2} x = 3 cos x + 3$ for $0 \leq x \leq 2 π$ , giving exact answers.
[5 marks]
Answer

Example 5

Solve $3 sin x = 2 tan x$ for $0 \leq x \leq 2 π$ , giving your answers to 3 significant figures where needed.

Hint Write

tan x = \frac{sin x}{cos x}

, multiply by

cos x

and bring everything to one side. Do not divide by

sin x

: factorise it out.

B5
Solve $5 sin x = 4 tan x$ for $0 \leq x \leq 2 π$ , giving your answers to 3 significant figures where needed.
[5 marks]
Answer

Practice

A-Level

C1
Sam solves $2 sin x cos x = sin x$ for $0 \leq x \leq 2 π$ . He divides both sides by $sin x$ to get $cos x = \frac{1}{2}$ , and gives $x = \frac{π}{3}, \frac{5 π}{3}$ . Explain his mistake and give every solution.
[5 marks]
Answer

C2
Solve $cos (2 x - \frac{π}{3}) = 0.4$ for $0 \leq x \leq π$ , giving your answers to 3 significant figures.
[5 marks]
Answer
C3
Zoe solves $tan x = 2$ for $0 \leq x \leq 2 π$ and gives $x = 63.4$ and $x = 243.4$ . Explain what she has done wrong and give the correct answers to 3 significant figures.
[4 marks]
Answer
C4
Solve $tan^{2} x = 3$ for $- π \leq x \leq π$ , giving exact answers.
[4 marks]
Answer
C5
Solve $2 sin (\frac{x}{2}) = 1$ for $0 \leq x \leq 4 π$ , giving exact answers.
[4 marks]
Answer

Exam-style questions

Edexcel exam style

D1
This question is about the equation $4 sin^{2} x + 8 cos x - 7 = 0$ .
(a)
Show that the equation can be written as $4 cos^{2} x - 8 cos x + 3 = 0$ .
[2 marks]
(b)
Hence solve $4 sin^{2} x + 8 cos x - 7 = 0$ for $0 \leq x \leq 2 π$ , giving exact answers.
[3 marks]
Answer
(c)
Hence solve $4 sin^{2} 2 θ + 8 cos 2 θ - 7 = 0$ for $0 \leq θ \leq 2 π$ , giving exact answers.
[4 marks]
Answer
Total for D1: 9 marks

D2
Leo solves $sin 2 x = \frac{1}{2}$ for $0 \leq x \leq 2 π$ . He writes: " $2 x = \frac{π}{6}$ or $\frac{5 π}{6}$ , so $x = \frac{π}{12}$ or $\frac{5 π}{12}$ ."
Leo is wrong.
(a)
Explain the mistake Leo has made.
[2 marks]
(b)
Solve $sin 2 x = \frac{1}{2}$ correctly for $0 \leq x \leq 2 π$ .
[3 marks]
Answer
(c)
Solve $sin 2 x = \frac{1}{2}$ for $- π \leq x \leq 0$ .
[2 marks]
Answer
Total for D2: 7 marks
D3
Solve $5 sin 3 θ = 2 cos 3 θ$ for $0 \leq θ < π$ , giving your answers to 3 significant figures.
[6 marks]
Answer
D4
The depth of water, $d$ metres, in a harbour is modelled by $d = 6 + 2.5 sin (\frac{π t}{6} - \frac{π}{3})$ , where $t$ is the time in hours after midnight and $0 \leq t \leq 12$ .
(a)
Find the greatest depth of water and the time it happens.
[2 marks]
Answer
(b)
Find the two times when the depth is $7$ m. Give each time to the nearest minute.
[4 marks]
Answer
(c)
The harbour master says that between midnight and noon the depth is at least $7$ m for more than a third of the time. Is she right? Show your working.
[3 marks]
Answer
Total for D4: 9 marks
D5
This question is about the equation $2 cos x = 3 tan x$ .
(a)
Show that the equation can be written as $2 sin^{2} x + 3 sin x - 2 = 0$ .
[3 marks]
(b)
Hence solve $2 cos x = 3 tan x$ for $0 \leq x \leq 2 π$ , giving exact answers. Explain why one factor gives no solutions.
[4 marks]
Answer
(c)
Hence solve $2 cos 2 θ = 3 tan 2 θ$ for $0 \leq θ \leq π$ .
[2 marks]
Answer
Total for D5: 9 marks
D6
This question is about the equation $2 sin^{2} x - cos x = k$ , where $k$ is a constant.
(a)
Show that the equation can be written as $2 cos^{2} x + cos x + k - 2 = 0$ .
[2 marks]
(b)
Given that $k = 1$ , solve the equation for $0 \leq x \leq 2 π$ , giving exact answers.
[4 marks]
Answer
(c)
Find the set of values of $k$ for which the equation has at least one solution.
[4 marks]
Answer
Total for D6: 10 marks

Extension

Stretch

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

E1
Solve $sin 3 x = cos 2 x$ for $0 \leq x \leq π$ , giving exact answers.
[9 marks]
Hint 1 · What to try
Write $cos 2 x$ as a sine: $cos 2 x = sin (\frac{π}{2} - 2 x)$ .
Hint 2 · The first line
If $sin A = sin B$ then either $A = B + 2 nπ$ or $A = π - B + 2 nπ$ , for a whole number $n$ .
Hint 3 · The full method
Each case gives a formula for $x$ with $n$ in it. Try $n = 0, 1, 2, \dots$ and $n = - 1, \dots$ and keep only the values between $0$ and $π$ .
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can find every solution of sin x = k, cos x = k or tan x = k in a given interval, in radians.
I can change the interval when the angle is 2x, x/2 or x + a, and change back at the end.
I can use sin²x + cos²x = 1 or tan x = sin x / cos x to make a quadratic in one ratio and solve it.
I can avoid losing solutions by factorising instead of dividing.

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