Trig Equations & General Solutions (Grade 11) — Worksheet & Memo

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Grade 11 · Trigonometry

Trig Equations & General Solutions (Grade 11)

EXAM-STYLE CLASS TEST

Marks

Duration

55 minutes

Questions

Name:

Class:

Date:

Mark

/ 34

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
The general solution of sin θ = 0,5 is …
(1)
A)θ = 30° + k(180°)
B)θ = 30° + k(360°) only
C)θ = 30° + k(360°) or θ = 150° + k(360°)
D)θ = 30° + k(360°) or θ = 330° + k(360°)
1.2
The general solution of cos θ = 0,5 is …
(1)
A)θ = 60° + k(360°) or θ = 120° + k(360°)
B)θ = 60° + k(360°) or θ = −60° + k(360°)
C)θ = 60° + k(180°)
D)θ = 60° + k(360°) only
1.3
The general solution of tan θ = 1 is …
(1)
A)θ = 45° only
B)θ = 45° + k(360°)
C)θ = 45° + k(180°)
D)θ = 45° or θ = 135° + k(360°)
1.4
Solve sin θ = −0,5 for 0° ≤ θ ≤ 360°.
(1)
A)210° and 330°
B)−30° only
C)30° and 150°
D)150° and 210°
1.5
Solve cos θ = −0,5 for 0° ≤ θ ≤ 360°.
(1)
A)120° and 240°
B)240° only
C)60° and 120°
D)60° and 300°
1.6
In which quadrants is tan θ POSITIVE?
(1)
A)the second and fourth
B)the first and third
C)the first and fourth
D)the first and second
1.7
Solve 2 sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.
(1)
A)30° and 330°
B)30° only
C)30° and 150°
D)60° and 120°
1.8
The general solution of sin 2θ = 0,5 is …
(1)
A)θ = 15° + k(360°) or θ = 75° + k(360°)
B)θ = 15° + k(180°) or θ = 75° + k(180°)
C)θ = 30° + k(360°) or θ = 150° + k(360°)
D)θ = 60° + k(180°)
1.9
Why is a GENERAL solution needed?
(1)
A)the reference angle is always negative
B)there is never more than one solution
C)the calculator gives a wrong answer
D)the trigonometric functions repeat, so there are infinitely many solutions
1.10
For sin θ = 0,8 the calculator gives 53,13°. The SECOND solution in [0° ; 360°] is …
(1)
A)36,87°
B)233,13°
C)306,87°
D)126,87°

Answer grid — circle your answers for Question 1

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[13 MARKS]

Determine the GENERAL SOLUTION of each equation, with k ∈ ℤ.

2.1
3 sin x − 1 = 0, correct to TWO decimal places
(5)
2.2
cos 2x = 0,5
(4)
2.3
tan(x − 30°) = 1
(4)

Question 3

[11 MARKS]

Answer the questions below.

3.1
Solve for x ∈ [0° ; 360°]: 2 cos²x − cos x − 1 = 0
(6)
3.2
Explain why a trigonometric equation has infinitely many solutions, but only a few in a given interval.
(3)
3.3
Explain why sin x = 1,5 has no solution.
(2)

TOTAL: 34 marks

This question paper consists of 3 questions.