ECAT Mathematics
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ECAT Mathematics
Chapter 14: Solutions of Trigonometric Equations
68 Questions
90 Minutes
Pass: 60%+
ECAT Prep
0
General solution of sinθ=0 is:
A
θ=2nπ
B
θ=nπ
C
θ=π/2+nπ
D
θ=π+nπ
1
General solution of cosθ=0 is:
A
θ=nπ
B
θ=π/2+nπ
C
θ=2nπ
D
θ=π+2nπ
2
General solution of tanθ=0 is:
A
θ=2nπ
B
θ=nπ
C
θ=π/2+nπ
D
θ=π+2nπ
3
General solution of sinθ=1 is:
A
θ=2nπ
B
θ=π/2+2nπ
C
θ=nπ
D
θ=π+2nπ
4
General solution of cosθ=1 is:
A
θ=π+2nπ
B
θ=nπ
C
θ=2nπ
D
θ=π/2+nπ
5
General solution of sinθ=sinα is:
A
θ=nπ+α
B
θ=2nπ±α
C
θ=nπ+(−1)ⁿα
D
θ=nπ−α
6
General solution of cosθ=cosα is:
A
θ=nπ+(−1)ⁿα
B
θ=nπ±α
C
θ=2nπ±α
D
θ=2nπ+α
7
General solution of tanθ=tanα is:
A
θ=2nπ+α
B
θ=nπ+α
C
θ=nπ±α
D
θ=(−1)ⁿα+nπ
8
Solutions of sinθ=1/2 in [0,2π):
A
π/6 only
B
π/6 and 5π/6
C
π/3 and 2π/3
D
π/4 and 3π/4
9
Solutions of cosθ=−1/2 in [0,2π):
A
π/3 and 5π/3
B
2π/3 and 4π/3
C
π/6 and 5π/6
D
2π/3 only
10
Solutions of tanθ=1 in [0,2π):
A
π/4 only
B
π/4 and 5π/4
C
π/4 and 3π/4
D
3π/4 only
11
sinθ=−1: solution in [0,2π):
A
π/2
B
π
C
3π/2
D
2π
12
cosθ=−1: solution in [0,2π):
A
0
B
π/2
C
π
D
3π/2
13
2sinθ=1 → sinθ=1/2. General solution:
A
θ=nπ+(−1)ⁿπ/6
B
θ=2nπ±π/6
C
θ=nπ+π/6
D
θ=2nπ+π/3
14
√3 tanθ=1 → tanθ=1/√3. General solution:
A
θ=nπ+π/6
B
θ=nπ+π/3
C
θ=nπ+π/4
D
θ=2nπ+π/6
15
2cosθ+√3=0 → cosθ=−√3/2. Solution in [0,2π):
A
π/6 and 5π/6
B
5π/6 and 7π/6
C
7π/6 and 11π/6
D
2π/3 and 4π/3
16
2sin²θ−sinθ−1=0. Factor: (2sinθ+1)(sinθ−1)=0. sinθ=?
A
1/2 or −1
B
−1/2 or 1
C
1 or −1
D
1/2 or 1
17
Solutions of sinθ=−1/2 in [0,2π):
A
7π/6 and 11π/6
B
5π/6 and 7π/6
C
π/6 and 5π/6
D
11π/6 only
18
Combined solutions of 2sin²θ−sinθ−1=0 in [0,2π):
A
π/2, 7π/6, 11π/6
B
π/2, π/6, 5π/6
C
π/2 only
D
7π/6, 11π/6 only
19
tan²θ=3 → tanθ=±√3. Solutions in [0,π):
A
π/3 only
B
2π/3 only
C
π/3 and 2π/3
D
π/4 and 3π/4
20
sin2θ=sinθ → 2sinθcosθ−sinθ=0 → sinθ(2cosθ−1)=0. Solutions:
A
sinθ=0 or cosθ=1/2
B
sinθ=1/2 or cosθ=0
C
sinθ=0 or cosθ=0
D
sinθ=1 or cosθ=1/2
21
sinθ=0 in [0,2π) gives:
A
θ=0,π
B
θ=π/2,3π/2
C
θ=0,π/2,π,3π/2
D
θ=0 only
22
cosθ=1/2 in [0,2π) gives:
A
π/3 and 5π/3
B
π/3 and π/6
C
2π/3 and 4π/3
D
π/4 and 7π/4
23
sin²θ=1/4 → sinθ=±1/2. Number of solutions in [0,2π):
A
2
B
4
C
3
D
1
24
The equation sinθ+cosθ=1. Divide by √2: sin(θ+π/4)=1/√2. Solutions in [0,2π):
A
θ=0 and π/2
B
θ=π/4 and 3π/4
C
θ=0 and 3π/2
D
θ=π/2 and 3π/2
25
sinθ+cosθ=√2. Rewrite as √2 sin(θ+π/4)=√2. sin(θ+π/4)=1. Solution:
A
θ+π/4=π/2 → θ=π/4
B
θ=3π/4
C
θ=π/2
D
θ=0
26
cos2θ=cosθ → 2cos²θ−cosθ−1=0 → (2cosθ+1)(cosθ−1)=0:
A
cosθ=−1/2 or 1
B
cosθ=1/2 or −1
C
cosθ=1 or −1
D
cosθ=1/2 or 1
27
cosθ=1 in [0,2π): θ=
A
0
B
π
C
2π
D
π/2
28
cosθ=−1/2 in [0,2π):
A
2π/3 and 4π/3
B
π/3 and 5π/3
C
π/6 and 5π/6
D
2π/3 only
29
2sinθcosθ=sinθ in [0,2π): factor sinθ(2cosθ−1)=0:
A
θ=0, π, π/3, 5π/3
B
θ=π/3, 5π/3
C
θ=0, π
D
θ=π/2, 3π/2
30
General solution of sinθ=−√3/2:
A
θ=nπ+(−1)ⁿ(π/3)
B
θ=nπ+(−1)ⁿ(−π/3)
C
θ=nπ+(−1)ⁿ(π/3)−π
D
Both B, since (−1)ⁿ(−π/3) gives the right values
31
tanθ=−1: solutions in [0,2π):
A
π/4 and 5π/4
B
3π/4 and 7π/4
C
π/4 and 3π/4
D
5π/4 only
32
Equation: tan2θ=tanθ → sin2θcosθ−cos2θsinθ=0 → sinθ=0 or... What is the simpler factoring?
A
tan2θ−tanθ=0; tanθ(2cosθ−1)=0
B
tan2θ=tanθ: general soln 2θ=nπ+θ → θ=nπ
C
cos2θ=0
D
sin2θ=0
33
Equation 4cos²θ−3=0 → cosθ=±√3/2. Number of solutions in [0,2π):
A
2
B
4
C
3
D
1
34
sin3θ=sinθ: using identity sin3θ−sinθ=2cos2θsinθ=0:
A
cos2θ=0 or sinθ=0
B
2cosθ=0
C
3θ=θ
D
sin2θ=0
35
sinθ=0: θ=nπ. cos2θ=0: 2θ=π/2+nπ → θ=
A
π/4+nπ/2
B
π/4+nπ
C
nπ
D
π/2+nπ
36
General solution of 2sin²θ+sinθ=0 → sinθ(2sinθ+1)=0:
A
θ=nπ or θ=nπ+(−1)ⁿ(7π/6)
B
θ=nπ or sinθ=−1/2
C
Only θ=nπ
D
θ=π/6 or π
37
cos²θ+cosθ=0 → cosθ(cosθ+1)=0:
A
cosθ=0 or −1
B
cosθ=1 or 0
C
cosθ=±1
D
cosθ=0 only
38
cosθ=0 → θ=π/2+nπ. cosθ=−1 → θ=
A
π+2nπ
B
π/2+nπ
C
2nπ
D
nπ
39
The equation 2cos²θ=1+sinθ (using cos²θ=1−sin²θ): 2−2sin²θ=1+sinθ → 2sin²θ+sinθ−1=0 → (2sinθ−1)(sinθ+1)=0:
A
sinθ=1/2 or −1
B
sinθ=−1/2 or 1
C
sinθ=1 or −1
D
sinθ=1/2 or 1
40
Solutions of sinθ=1/2 in [0,2π): π/6 and 5π/6. Solution of sinθ=−1:
A
3π/2
B
π/2
C
π
D
2π
41
All solutions of 2cos²θ=1+sinθ in [0,2π):
A
π/6, 5π/6, 3π/2
B
π/3, 2π/3, π/2
C
π/4, 3π/4, 3π/2
D
π/6, 5π/6, π/2
42
Number of solutions of sinx=x/100 in (−π,π):
A
1
B
10
C
63
D
infinite
43
tan²θ+secθ=1 (using sec²θ=1+tan²θ): sec²θ−1+secθ=1→sec²θ+secθ−2=0→(secθ+2)(secθ−1)=0:
A
secθ=1 or −2
B
secθ=2 or −1
C
secθ=1 or 2
D
secθ=−1 or −2
44
secθ=1→cosθ=1→θ=2nπ. secθ=−2→cosθ=−1/2→θ=
A
2π/3+2nπ or 4π/3+2nπ
B
π/3+2nπ
C
π/6+nπ
D
2nπ
45
Equation sinθ=2 has:
A
2 solutions
B
1 solution
C
No real solution
D
Infinite solutions
46
Equation cosθ=−2 has:
A
2 solutions
B
1 solution
C
No real solution
D
Infinite solutions
47
Equation sinθ+sin2θ+sin3θ=0: using sum-to-product sin3θ+sinθ=2sin2θcosθ, so 2sin2θcosθ+sin2θ=0→sin2θ(2cosθ+1)=0:
A
sin2θ=0 or cosθ=−1/2
B
sin2θ=0 or cosθ=1/2
C
sinθ=0 or sin2θ=0
D
sin2θ=0 only
48
sin2θ=0 → 2θ=nπ → θ=nπ/2. In [0,π):
A
θ=0,π/2,π
B
θ=0,π/2
C
θ=π/4,3π/4
D
θ=0,π/3,2π/3
49
cosθ=−1/2 in [0,π):
A
π/3
B
2π/3
C
π/2
D
π
50
Equation cos2θ+3cosθ+2=0 (using cos2θ=2cos²θ−1): 2cos²θ+3cosθ+1=0 → (2cosθ+1)(cosθ+1)=0:
A
cosθ=−1/2 or −1
B
cosθ=1/2 or 1
C
cosθ=−1 or 1
D
cosθ=−1/2 or 1
51
General solutions of equations with sum of trig functions are found by:
A
Only numerical methods
B
Algebraic manipulation, identities, and standard solution forms
C
Graphical methods only
D
Calculus only
52
Equation sin(x+π/4)=1/√2: x+π/4=π/4+2nπ or x+π/4=3π/4+2nπ. Solutions:
A
x=0+2nπ or x=π/2+2nπ
B
x=π/4 or x=3π/4
C
x=0 or x=π
D
x=π/2 only
53
Trigonometric equation may have:
A
Only finite solutions
B
Only infinite solutions
C
Finite or infinite based on interval specified
D
No solutions
54
For equations of form Rcosθ=a, Rsinθ=b, written as Rsin(θ+φ)=c:
A
R=√(a²+b²)
B
R=a+b
C
R=ab
D
R=a/b
55
4sin²θ=3 → sinθ=±√3/2. Number of solutions in [0,2π):
A
2
B
4
C
3
D
1
56
3tan²θ=1 → tanθ=±1/√3. Number of solutions in [0,π):
A
1
B
2
C
3
D
4
57
The equation cos²x−sinx−1/4=0 becomes (1−sin²x)−sinx−1/4=0→4sin²x+4sinx−3=0→(2sinx+3)(2sinx−1)=0:
A
sinx=−3/2 or 1/2
B
sinx=3/2 or −1/2
C
sinx=1/2 only
D
sinx=−3/2 only
58
Solutions of sinx=1/2 in [0,2π):
A
π/6 and 5π/6
B
π/3 and 2π/3
C
π/4 and 3π/4
D
π/6 only
59
sinθ=sin(π/7) → general solution: θ=
A
nπ+(−1)ⁿ(π/7)
B
nπ+π/7
C
2nπ±π/7
D
nπ±π/7
60
cosθ=cos(π/5) → θ=
A
nπ±π/5
B
2nπ±π/5
C
nπ+(−1)ⁿπ/5
D
nπ+π/5
61
tanθ=tan(π/8) → θ=
A
nπ+π/8
B
2nπ+π/8
C
nπ±π/8
D
(−1)ⁿπ/8
62
Simultaneous equations sinθ=1/2 AND cosθ=√3/2 have solution:
A
θ=π/6
B
θ=5π/6
C
θ=π/3
D
θ=5π/3
63
Equation sin²θ=cos²θ → tan²θ=1 → tanθ=±1. Solutions in [0,2π):
A
π/4,3π/4,5π/4,7π/4
B
π/4,5π/4
C
π/4,3π/4
D
3π/4,7π/4
64
sinθ−sin3θ=cos2θ: use sinθ−sin3θ=−2cos2θsinθ: −2cos2θsinθ=cos2θ→cos2θ(2sinθ+1)=0:
A
cos2θ=0 or sinθ=−1/2
B
sin2θ=0 or cosθ=−1/2
C
cos2θ=0 only
D
sinθ=−1/2 only
65
When solving trig equations, extraneous solutions arise when:
A
Using quadratic formula
B
Squaring both sides
C
Factoring
D
Using identities directly
66
sinx+cosx=0 → tanx=−1 → x=nπ−π/4. In [0,2π):
A
3π/4 and 7π/4
B
π/4 and 5π/4
C
π/4 and 3π/4
D
5π/4 and 7π/4
67
2sin²x−sinx=0→sinx(2sinx−1)=0: solutions in [0,2π):
A
0,π/6,5π/6,π
B
0,π,π/6,5π/6
C
π/6,5π/6
D
0,π
0
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