Sec 3 A-Math Trigonometric equations worksheet with answers

A free Sec 3 A-Math worksheet on trigonometric equations: 9 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 A-Math worksheet: Trigonometric equations

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Without using a calculator, find the principal value, in degrees, of

\sin^{-1}\left(-\dfrac{1}{2}\right)

Answer[2]

The temperature,

T

°C, in a greenhouse

t

hours after midnight is given by

T = 25 - 6\cos (15t)^\circ

. Find the length of time in one complete cycle,

0 \leq t \leq 24

, for which the temperature is more than

28

°C.

Answerhours[3]

Solve

\sin x = -\dfrac{1}{2}

for

0^\circ \leq x \leq 360^\circ

Answer[3]

Solve

\cos (x + 30^\circ) = -\dfrac{\sqrt{3}}{2}

for

0^\circ \leq x \leq 360^\circ

Answer[3]

Solve

\operatorname{cosec} x = -2

for

0^\circ \leq x \leq 360^\circ

Answer[3]

Solve

\cos 4x = -\dfrac{\sqrt{3}}{2}

for

0^\circ \leq x \leq 360^\circ

Answer[3]

Section B

(a)Show that the equation

{12\cos x + 2\sec x = 11}

can be written as

{12\cos^{2} x - 11\cos x + 2 = 0}

[2]

(b)Hence solve the equation for

0^\circ \leq x \leq 360^\circ

, giving your answers correct to 1 decimal place where necessary.

Answer (b)°[3]

(c)Hence solve the equation

{12\cos 2y + 2\sec 2y = 11}

for

0^\circ \leq y \leq 180^\circ

, giving your answers correct to 1 decimal place where necessary.

Answer (c)°[2]

(a)Show that the equation

{9\sin x - 4\operatorname{cosec} x = 9}

can be written as

{9\sin^{2} x - 9\sin x - 4 = 0}

[2]

(b)Hence solve the equation for

0^\circ \leq x \leq 360^\circ

, giving your answers correct to 1 decimal place where necessary.

Answer (b)°[3]

(c)Hence solve the equation

{9\sin 2y - 4\operatorname{cosec} 2y = 9}

for

0^\circ \leq y \leq 180^\circ

, giving your answers correct to 1 decimal place where necessary.

Answer (c)°[2]

In this question, angles are in radians.

(a)Find the exact solutions of the equation

{\tan\left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{\sqrt{3}}}

for

0 \leq x \leq 2\pi

Answer (a)[3]

(b)Solve the equation

{3\sin x = -\cos x}

for

0 \leq x \leq 2\pi

, giving your answers correct to 2 decimal places.

Answer (b)[3]

(c)Solve the equation

{2\sin^{2} x = \sin x}

for

0 \leq x \leq 2\pi

, giving exact answers.

Answer (c)[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

\sin^{-1}\left(-\dfrac{1}{2}\right) = -30^\circ

[2]

\text{Basic angle} = 30^\circ

The principal value of

\sin^{-1}

lies in

-90^\circ

90^\circ

\sin^{-1}\left(-\dfrac{1}{2}\right) = -30^\circ

Question 2

Answer:

\text{Time} = 8

hours[3]

25 - 6\cos (15t)^\circ = 28

-6\cos (15t)^\circ = 28 - 25

\cos (15t)^\circ = - \dfrac{1}{2}

\alpha = \cos^{-1}\left(\dfrac{1}{2}\right)

= 60^\circ

\cos u

is negative in quadrants 2 and 3.

u = 180^\circ - 60^\circ

= 120^\circ

u = 180^\circ + 60^\circ

= 240^\circ

t = \dfrac{u}{15}

t = 8

t = 16

t = 12

T = 31

, so

T > 28

between these times.

\text{Between them} = 16 - 8

= 8

hours

\text{Time} = 8

hours

Question 3

Answer:

= 330^\circ

[3]

\alpha = \sin^{-1}\left(\dfrac{1}{2}\right)

= 30^\circ

\sin x

is negative in quadrants 3 and 4.

x = 180^\circ + 30^\circ

= 210^\circ

x = 360^\circ - 30^\circ

= 330^\circ

Question 4

Answer:

x = 180^\circ

[3]

\text{Lowest } u = 30^\circ

\text{Highest } u = 390^\circ

\alpha = \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)

= 30^\circ

\cos u

is negative in quadrants 2 and 3.

u = 180^\circ - 30^\circ

= 150^\circ

u = 180^\circ + 30^\circ

= 210^\circ

x = u - 30^\circ

x = 120^\circ

x = 180^\circ

Question 5

Answer:

= 330^\circ

[3]

\sin x = -\dfrac{1}{2}

\alpha = \sin^{-1}\left(\dfrac{1}{2}\right)

= 30^\circ

\sin x

is negative in quadrants 3 and 4.

x = 180^\circ + 30^\circ

= 210^\circ

x = 360^\circ - 30^\circ

= 330^\circ

Question 6

Answer:

x = 322.5^\circ

[3]

\text{Lowest } u = 0^\circ

\text{Highest } u = 1440^\circ

\alpha = \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)

= 30^\circ

\cos u

is negative in quadrants 2 and 3.

u = 180^\circ - 30^\circ

= 150^\circ

u = 180^\circ + 30^\circ

= 210^\circ

u = 180^\circ - 30^\circ + 360^\circ

= 510^\circ

u = 180^\circ + 30^\circ + 360^\circ

= 570^\circ

u = 180^\circ - 30^\circ + 720^\circ

= 870^\circ

u = 180^\circ + 30^\circ + 720^\circ

= 930^\circ

u = 180^\circ - 30^\circ + 1080^\circ

= 1230^\circ

u = 180^\circ + 30^\circ + 1080^\circ

= 1290^\circ

x = \dfrac{u}{4}

x = 37.5^\circ

x = 52.5^\circ

x = 127.5^\circ

x = 142.5^\circ

x = 217.5^\circ

x = 232.5^\circ

x = 307.5^\circ

x = 322.5^\circ

Question 7

(a)[2]

12\cos x + \dfrac{2}{\cos x} = 11

12\cos^{2} x + 2 = 11\cos x

12\cos^{2} x - 11\cos x + 2 = 0

(shown)

(b)Answer:

x = 48.2^\circ

x = 75.5^\circ

x = 284.5^\circ

x = 311.8^\circ

°[3]

(3\cos x - 2)(4\cos x - 1) = 0

\cos x = \dfrac{2}{3}

\cos x = \dfrac{1}{4}

x = 48.2^\circ

x = 75.5^\circ

x = 284.5^\circ

x = 311.8^\circ

(c)Answer:

y = 24.1^\circ

y = 37.8^\circ

y = 142.2^\circ

y = 155.9^\circ

°[2]

Let

x = 2y

0^\circ \leq 2y \leq 360^\circ

2y = 48.2^\circ, 75.5^\circ, 284.5^\circ, 311.8^\circ

y = 24.1^\circ

y = 37.8^\circ

y = 142.2^\circ

y = 155.9^\circ

Question 8

(a)[2]

9\sin x - \dfrac{4}{\sin x} = 9

9\sin^{2} x - 4 = 9\sin x

9\sin^{2} x - 9\sin x - 4 = 0

(shown)

(b)Answer:

x = 199.5^\circ

x = 340.5^\circ

°[3]

(3\sin x + 1)(3\sin x - 4) = 0

\sin x = -\dfrac{1}{3}

\sin x = \dfrac{4}{3}

\sin x = \dfrac{4}{3}

has no solution, as

-1 \leq \sin x \leq 1

x = 199.5^\circ

x = 340.5^\circ

(c)Answer:

y = 99.7^\circ

y = 170.3^\circ

°[2]

Let

x = 2y

0^\circ \leq 2y \leq 360^\circ

2y = 199.5^\circ, 340.5^\circ

y = 99.7^\circ

y = 170.3^\circ

Question 9

(a)Answer:

\dfrac{\pi}{2}

\dfrac{3\pi}{2}

[3]

Basic angle

= \dfrac{\pi}{6}

\dfrac{-\pi}{3} \leq x - \dfrac{\pi}{3} \leq \dfrac{5\pi}{3}

x - \dfrac{\pi}{3} = \dfrac{\pi}{6}, \dfrac{7\pi}{6}

x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}

(b)Answer:

2.82

5.96

[3]

\tan x = -\dfrac{1}{3}

Basic angle

= \tan^{-1} \dfrac{1}{3}

= 0.3218

\tan x < 0

x

in quadrants 2 and 4

x = 2.82, 5.96

(c)Answer:

0

\pi

2\pi

\dfrac{\pi}{6}

\dfrac{5\pi}{6}

[3]

2\sin^{2} x - \sin x = 0

\sin x(2\sin x - 1) = 0

\sin x = 0

\sin x = \dfrac{1}{2}

\sin x = 0

x = 0, \pi, 2\pi

\sin x = \dfrac{1}{2}

x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}

(Dividing by

\sin x

would lose the solutions of

\sin x = 0

Practice questions by skill

Sec 3 A-Math Trigonometric equations worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Trigonometric equations notes. Or practise it online with step-by-step help when you get one wrong.

Section B

(a)Show that the equation

{12\cos x + 2\sec x = 11}

can be written as

{12\cos^{2} x - 11\cos x + 2 = 0}

[2]

(b)Hence solve the equation for

0^\circ \leq x \leq 360^\circ

, giving your answers correct to 1 decimal place where necessary.

Answer (b)°[3]

(c)Hence solve the equation

{12\cos 2y + 2\sec 2y = 11}

for

0^\circ \leq y \leq 180^\circ

, giving your answers correct to 1 decimal place where necessary.

Answer (c)°[2]

Answers and worked solutions

Question 1

Answer:

\sin^{-1}\left(-\dfrac{1}{2}\right) = -30^\circ

[2]

\text{Basic angle} = 30^\circ

The principal value of

\sin^{-1}

lies in

-90^\circ

90^\circ

\sin^{-1}\left(-\dfrac{1}{2}\right) = -30^\circ

Question 2

Answer:

\text{Time} = 8

hours[3]

25 - 6\cos (15t)^\circ = 28

-6\cos (15t)^\circ = 28 - 25

\cos (15t)^\circ = - \dfrac{1}{2}

\alpha = \cos^{-1}\left(\dfrac{1}{2}\right)

= 60^\circ

\cos u

is negative in quadrants 2 and 3.

u = 180^\circ - 60^\circ

= 120^\circ

u = 180^\circ + 60^\circ

= 240^\circ

t = \dfrac{u}{15}

t = 8

t = 16

t = 12

T = 31

, so

T > 28

between these times.

\text{Between them} = 16 - 8

= 8

hours

\text{Time} = 8

hours

Question 3

Answer:

= 330^\circ

[3]

\alpha = \sin^{-1}\left(\dfrac{1}{2}\right)

= 30^\circ

\sin x

is negative in quadrants 3 and 4.

x = 180^\circ + 30^\circ

= 210^\circ

x = 360^\circ - 30^\circ

= 330^\circ

Question 4

Answer:

x = 180^\circ

[3]

\text{Lowest } u = 30^\circ

\text{Highest } u = 390^\circ

\alpha = \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)

= 30^\circ

\cos u

is negative in quadrants 2 and 3.

u = 180^\circ - 30^\circ

= 150^\circ

u = 180^\circ + 30^\circ

= 210^\circ

x = u - 30^\circ

x = 120^\circ

x = 180^\circ

Question 5

Answer:

= 330^\circ

[3]

\sin x = -\dfrac{1}{2}

\alpha = \sin^{-1}\left(\dfrac{1}{2}\right)

= 30^\circ

\sin x

is negative in quadrants 3 and 4.

x = 180^\circ + 30^\circ

= 210^\circ

x = 360^\circ - 30^\circ

= 330^\circ

Question 6

Answer:

x = 322.5^\circ

[3]

\text{Lowest } u = 0^\circ

\text{Highest } u = 1440^\circ

\alpha = \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)

= 30^\circ

\cos u

is negative in quadrants 2 and 3.

u = 180^\circ - 30^\circ

= 150^\circ

u = 180^\circ + 30^\circ

= 210^\circ

u = 180^\circ - 30^\circ + 360^\circ

= 510^\circ

u = 180^\circ + 30^\circ + 360^\circ

= 570^\circ

u = 180^\circ - 30^\circ + 720^\circ

= 870^\circ

u = 180^\circ + 30^\circ + 720^\circ

= 930^\circ

u = 180^\circ - 30^\circ + 1080^\circ

= 1230^\circ

u = 180^\circ + 30^\circ + 1080^\circ

= 1290^\circ

x = \dfrac{u}{4}

x = 37.5^\circ

x = 52.5^\circ

x = 127.5^\circ

x = 142.5^\circ

x = 217.5^\circ

x = 232.5^\circ

x = 307.5^\circ

x = 322.5^\circ

Question 7

(a)[2]

12\cos x + \dfrac{2}{\cos x} = 11

12\cos^{2} x + 2 = 11\cos x

12\cos^{2} x - 11\cos x + 2 = 0

(shown)

(b)Answer:

x = 48.2^\circ

x = 75.5^\circ

x = 284.5^\circ

x = 311.8^\circ

°[3]

(3\cos x - 2)(4\cos x - 1) = 0

\cos x = \dfrac{2}{3}

\cos x = \dfrac{1}{4}

x = 48.2^\circ

x = 75.5^\circ

x = 284.5^\circ

x = 311.8^\circ

(c)Answer:

y = 24.1^\circ

y = 37.8^\circ

y = 142.2^\circ

y = 155.9^\circ

°[2]

Let

x = 2y

0^\circ \leq 2y \leq 360^\circ

2y = 48.2^\circ, 75.5^\circ, 284.5^\circ, 311.8^\circ

y = 24.1^\circ

y = 37.8^\circ

y = 142.2^\circ

y = 155.9^\circ

Question 8

(a)[2]

9\sin x - \dfrac{4}{\sin x} = 9

9\sin^{2} x - 4 = 9\sin x

9\sin^{2} x - 9\sin x - 4 = 0

(shown)

(b)Answer:

x = 199.5^\circ

x = 340.5^\circ

°[3]

(3\sin x + 1)(3\sin x - 4) = 0

\sin x = -\dfrac{1}{3}

\sin x = \dfrac{4}{3}

\sin x = \dfrac{4}{3}

has no solution, as

-1 \leq \sin x \leq 1

x = 199.5^\circ

x = 340.5^\circ

(c)Answer:

y = 99.7^\circ

y = 170.3^\circ

°[2]

Let

x = 2y

0^\circ \leq 2y \leq 360^\circ

2y = 199.5^\circ, 340.5^\circ

y = 99.7^\circ

y = 170.3^\circ

Question 9

(a)Answer:

\dfrac{\pi}{2}

\dfrac{3\pi}{2}

[3]

Basic angle

= \dfrac{\pi}{6}

\dfrac{-\pi}{3} \leq x - \dfrac{\pi}{3} \leq \dfrac{5\pi}{3}

x - \dfrac{\pi}{3} = \dfrac{\pi}{6}, \dfrac{7\pi}{6}

x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}

(b)Answer:

2.82

5.96

[3]

\tan x = -\dfrac{1}{3}

Basic angle

= \tan^{-1} \dfrac{1}{3}

= 0.3218

\tan x < 0

x

in quadrants 2 and 4

x = 2.82, 5.96

(c)Answer:

0

\pi

2\pi

\dfrac{\pi}{6}

\dfrac{5\pi}{6}

[3]

2\sin^{2} x - \sin x = 0

\sin x(2\sin x - 1) = 0

\sin x = 0

\sin x = \dfrac{1}{2}

\sin x = 0

x = 0, \pi, 2\pi

\sin x = \dfrac{1}{2}

x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}

(Dividing by

\sin x

would lose the solutions of

\sin x = 0