Download Section 8.3 – Trigonometric Equations

99
Section 8.3 – Trigonometric Equations
Objective 1:
Solve Equations Involving One Trigonometric Function.
In this section and the next, we will explore how to solving equations
involving trigonometric functions. Many times, an equation with a
trigonometric function will have an infinite number of solutions. Thus, we
will need to write the general form of the solution.
Find the a) general solution as well as b) eight particular solutions:
Ex. 1
3
2
cos(x) =
Solution:
3
If we examine the graph of y = cos(x) and y =
, we see that there
2
are an€infinite number of points the two curves intersect:
1.5
1
€
y=
0.5
3
2
y = cos(x)
0
– 4π
– 3π
– 2π
–π
π
-0.5
2π
3π
4π
€
-1
-1.5
In the interval [0, 2π), the cosine function is positive in the first and
fourth quadrants. In the first quadrant, x = cos – 1(
is also our reference angle, then 2π –
π
6
=
11π
6
3
2
)=
π
6
. Since
is the angle in the
fourth quadrant. The cosine function is periodic with period 2π, so
€ our general
€
€
these solutions will repeat every 2π. Thus,
solution
is
π
11π
a)
{x | x =
+ 2kπ or x = € +€2kπ, k is an integer}
b)
6
To find eight solutions, we can pick k = – 1, 0, 1, 2
k=–1
k=0
π
11π
π
π
+
2(–
1)π
=
–
+
2(0)π
=
€
€
6
11π
6
€
€
6
+ 2(– 1)π = –
€
€
6
π
6
€
€
6
11π
6
6
+ 2(0)π =
€
€
11π
6
π
6
100
k=1
π
+ 2(1)π =
6
11π
6
k=2
π
+ 2(2)π =
11π
So, {–
,
,
€
€ 6
solutions to the equation.
€
€
€
€
Ex. 2
+
13π
6
23π
2(1)π =
6
11π
π
π
,– ,
6
6
6
25π
6
6
11π
35π
+ 2(2)π =
6
6
13π 23π 25π 35π
,
,
,
}
6
6
6
€6
are eight
€
3
€ sin(θ)€= –€ 2€
€
€
€
€
Solution:
In the interval [0, 2π), the sine function is negative in the third and
fourth quadrants.
In the first quadrant, θ = sin – 1(
€
3
2
)=
π
3
. Thus,
is our reference angle. The angle we need in the third quadrant is
π
4π
π+
=
and the angle we need in the fourth quadrant is 2π –
3
=
€
3
€
€
3
To find eight solutions, we can pick k = – 1, 0, 1, 2
k=–1
k=0
4π
2π
4π
4π
+ 2(– 1)π =€–
+ 2(0)π =
€
3
5π
3
€
π
3
€ so these
€ solutions
. The sine function is periodic with€period 2π,
will repeat every 2π. Thus, our general solution is
4π
5π
€
a)
{x | x =
+ 2kπ or x =
+ 2kπ, k is an integer} €
b)
€
5π
3
3
π
3
+ 2(– 1)π = –
k=1
4π
€ =
+ 2(1)π
3
π
3
10π €
3
3
5π
11π
€ =
€
+ 2(1)π
3
3
2π
π 4π 5π
So, {–
,– ,
,
3
3 € 3
€ 3
3
5π
3
+ 2(0)π =
k=2
4π €
+ 2(2)π =
,
3
5π €
+ 2(2)π =
3
10π 11π 16π
,
,
,
3
3€
3
solutions to the equation.
€
€
€
2
Ex. 3
5cot (x) – 3 = 2
€
€
€
Solution: € € € €
First, we need to solve for cot(x):
5cot2(x) – 3 = 2
5cot2(x) = 5
cot2(x) = 1 or cot(x) = ± 1
€
€
3
5π
3
16π
3
17π
3
17π
}
3
are eight
101
The cotangent function is positive in quadrant I and negative in
π
quadrant II. Since cot(x) = 1 when tan(x) = 1, then x =
is
4
the reference angle. In quadrant II, the angle we need is π –
=
π
4
3π
4
. Since the cotangent function is periodic with period of π, then
€
the solutions will repeat every multiple of π.
π
3π
€ Since the
a)
{x | x =
+ nπ or x =
+ nπ, n is an integer}.
4
€
4
consecutive angles differ by
kπ
π
π
2
, we can write the general
solution as: {x | x =
+
, k is an integer}.
4
2
€
€
3π
π
π 3π 5π 7π 9π 11π
b)
Eight solutions are {–
,– , ,
,
,
,
,
}.
4
4
4
4
4
4
4
4
€
When solving a trigonometric equation where the argument is a multiple of
€
€
the variable, we will need to use the general solution for argument and then
solve for variable. We will€then €
need
through
€ to€run €
€ values
€ €of k until we find
all the angles within a specified interval.
Solve the following for angles in [0, 2π).
Ex. 4
2cos(4θ) – 1 = 0
Solution:
First, solve for cos(4θ):
2cos(4θ) – 1 = 0
2cos(4θ) = 1
cos(4θ) =
1
2
(let x = 4θ)
Thus, cos(x) =
1
2
π
3
when x =
or x = 2π –
π
3
=
5π
3
. Since the period
of cosine is 2π, then the general solution will have the form of:
€ x = 4θ = π + 2kπ or 5π + 2kπ
3
3
Now, solve
for θ by€dividing by four:
€
€
θ=
π
12
+
kπ
2
or
5π
12
+
€
kπ
2
€ to find all€the angles in [0, 2π).
We need
π
−1π
5π
5π
k=–1
θ=
+
=–
or
+
12
π
12
π
12
2
0π
+€
2
1π
+
€2
12
π
12
7π
12
€
5π
12
5π
€12
−1π
=
12
2
0π
5π
+
=
2
12
1π
11π
+
=
2 € 12
k€= 0 €
θ€=
k=1
€
θ=
€
€
€
€
€
€
€
€
€
€
€
€
€
=
=
or
or
–
π
12
No
Yes
Yes
102
π
+
12
π
k=3
θ=
+
12
π
k=4
θ=
+
12
€
€
€
π
The solution is { ,
€
€ 12€
k=2
Ex. 5
θ=
2π
13π
=
or
2
12
3π
19π
=
or
2
12
4π
25π
=
or
2
€ 12 €
5π 7π 11π
,
,
,
12 12
€ 12€
2π
2
3π
2
4π
2
17π
12
17π
12
23π
=
12
29π
=
12
19π 23π
,
,
12
12
=
Yes
Yes
No
}
€
€ + 3€ 3 = 3 € 3 €
€
4sin(3θ)
Solution:
€ € € €
€
€
€
€
We first need to solve for sin(3θ):
4sin(3θ) + 3 3 = 3 3
€
€
4sin(3θ) = 0
sin(3θ) = 0
(let x = 3θ)
Thus, sin(x) = 0 when x = 0 or π. Since the period of sine is 2π, then
€
€
the general solution will have the form of:
x = 3θ = 0 + 2kπ or π + 2kπ
Since consecutive angles differ by π, then we can state the general
solution as:
x = kπ
Now, solve for θ by dividing by 3:
θ=
kπ
3
We need to find all the angles in [0, 2π).
−1π
π
=–
No
3
3
1π
π
k=1
θ=
=
Yes
3
3
€
3π
k=3
θ=
= π Yes
€
€3
5π
k=5
θ=
Yes
€
€ 3
π 2π
4π
The solution is {0, ,
, π,
3
3
3
€
k=–1
Ex. 6
5π
+
12
5π
+
12
5π
+
12 €
13π
,
12 €
θ=
€
tan(2θ
+
π
2
k=0
θ=
k=2
θ=
k=4
k=6
,
5π
3
}.
3
)=–
θ=
€
θ=
€
0π
3
2π
3
4π
3
6π
3
= 0 Yes
Yes
Yes
= 2π No
€
€
€ €
Solution: € €
π
π
π
Let x = 2θ + . Since tan(x) = 3 when x = , then
is our
2
3
3
€ The tangent function is negative in quadrant II, so
reference
€ angle.
the angle is π –
π
3
=
2π
3
. Since the period of tangent is π, then the
€ the form of:
€ solution will have
€
general
€
€
€
103
x = 2θ +
π
2
=
Now, solve for θ:
π
2π
=
2
3
2π
π
€
€
2θ =
–
3
2
π
2θ =
+ kπ
€ 6
π
kπ
θ=
+
12
2
€
2θ +
€
2π
3
+ kπ
+ kπ
+ kπ
€
We need to find all the angles in [0, 2π).
π
−1π
5π
k=–1€ θ=
+
=–
No
k=0
k=1 €
k=3
€
The
€
2
12
1π
7π
€
θ=
+
=
Yes
2
12
3π
19π
θ=
+
=
Yes
2
12
€
€
π
7π 13π 19π
solution is { ,
,
,
12
12
12
12
€
€
€ 2:€
Objective
12
π
12
π
12
k=2
k=4
}.
π
+
12
π
θ=
+
12
π
θ=
+
12
€
€
θ=
€
€
€
0π
π
=
Yes
2
12
2π
13π
=
Yes
2
12
4π
25π
=
No
2
12
€
€
€ with
€ a Calculator.
€
Solving a Trigonometric Equation
€ € €
€
Solve the following
for angles
in [0, 2π).
Ex. 7
cos(x) = 0.6
Solution:
Here, we will need to use our calculator to find the value for x:
cos(x) = 0.6
x = cos – 1(0.6) = 0.927295218… ≈ 0.9273
This angle is in the first quadrant. However, the cosine is also positive
in the fourth quadrant. Since θR ≈ 0.9273, then the angle in the fourth
quadrant is 2π – 0.927295218… ≈ 5.3559
Thus, the solution is {≈ 0.9273, ≈ 5.3559}
Ex. 8
tan(x) = – 2
Solution:
Here, we will need to use our calculator to find the value for x:
tan(x) = – 2 which means x = tan – 1(– 2) = – 1.1071487…
This angle is not in [0, 2π). In that interval, the tangent function is
negative in quadrant II, and IV. Since the inverse tangent function
gave us the opposite of the reference angle, then θR = 1.1071487…
Thus, in the second quadrant, the angle is π – 1.1071487…
104
= 2.034443… ≈ 2.0344. Similarly, in quadrant IV, the angle is
2π – 1.1071487… = 5.1760365… ≈ 5.1760
Hence, the solution is {≈ 2.0344, ≈ 5.1760}.
Ex. 9
Due to bad weather, a plane in a holding pattern around the
Dallas airport. The distance d in miles the plane is from the airport at
time t minutes is given by d(t) = 80sin(0.55t) + 130.
a)
When the plane enters the holding pattern, t = 0, how far is it
from the airport?
b)
During the first 20 minutes after the plane enters the holding
pattern, what time(s) t will the plane be exactly 80 miles
from the airport?
Solution:
a)
d(0) = 80sin(0.55(0)) + 130 = 80sin(0) + 130 = 0 + 130 = 130
The plane was 130 miles from the airport.
b)
Set d(t) = 80 and solve:
80sin(0.55t) + 130 = 80
80sin(0.55t) = – 50
sin(0.55t) = – 0.625
0.55t = sin – 1(– 0.625) = – 0.67513153…
Thus, the reference angle is 0.67513153…
The sine is negative in quadrant III and IV, so the angles are
π + 0.67513153… = 3.8167241… and
2π – 0.67513153… = 5.6080537…
Thus, 0.55t = 3.8167241… + 2kπ or 5.6080537… + 2kπ, k is an
integer. Solving for t yields:
40kπ
40kπ
t = 6.9394985… +
or 10.196461… +
If k = 0, then
t = 6.9394985… +
11
40(0)π
11
11
or 10.196461… +
t ≈ 6.9395€minutes or 10.1965 minutes
€
If k = 1, then
40(1)π
t = 6.9394985…
+
or 10.196461…
+
€
€
11
40(0)π
11
40(1)π
11
t = 6.9394985… + 11.423973… or 10.196461… + 11.423973…
t = 18.363471… or 21.620434…
t ≈ 18.3635 minutes or 21.6204 minutes
€
The three times€that are within the first twenty
minutes are
{≈ 6.9395 minutes, ≈ 10.1965 minutes, ≈ 18.3635 minutes}
105
Objective 3:
Solving a Trigonometric Equation in Quadratic Form
In solving trigonometric equation in quadratic form, we will have to use an
identity and/or factor before we can get a series of linear equations
involving one trigonometric function to solve. It will also be important to
make a note of any values that make the original equation undefined.
Solve for all values in [0, 2π):
Ex. 10
2cos2(θ) + cos(θ) = 1
Solution:
2cos2(θ) + cos(θ) = 1
(get zero on one side)
2
2cos (θ) + cos(θ) – 1 = 0
Think of 2x2 + x – 1 = (2x – 1)(x + 1), so
2cos2(θ) + cos(θ) – 1 = (2cos(θ) – 1)(cos(θ) + 1)
Hence, (2cos(θ) – 1)(cos(θ) + 1) = 0
(solve)
2cos(θ) – 1 = 0 or
cos(θ) + 1 = 0
cos(θ) =
θ=
π
3
1
2
or
or 2π –
π
3
The solution is {
€
5π
3
=
π
3
cos(θ) = – 1
, π,
or
5π
3
θ=π
}.
Ex.
3sin2€
(θ) – 8sin(θ) – 11 = 0
€ 11
€
Solution:
€ – 11 = 0
3sin2(θ)€– 8sin(θ)
Think of 3x2 – 8x – 11 = (3x – 11)(x + 1), so
3sin2(θ) – 8sin(θ) – 11 = (3sin(θ) – 11)(sin(θ) + 1)
Hence, (3sin(θ) – 11)(sin(θ) + 1) = 0
(solve)
(3sin(θ) – 11) = 0
or
(sin(θ) + 1) = 0
sin(θ) =
11
3
or
No solution
θ=
The solution is {
€
Objective 4:
sin(θ) = – 1
3π
2
3π
2
}.
€
Solving Trigonometric
Equations Using Identities.
€ 2(θ) = cot(θ) + 1
Ex. 12
csc
Solution:
106
The cosecant and cotangent function is undefined when the sin(θ) = 0
or θ = 0 or π. Thus, our restrictions are θ ≠ 0 or π.
csc2(θ) = cot(θ) + 1
(csc2(θ) = cot2(θ) + 1)
cot2(θ) + 1 = cot(θ) + 1
(subtract cot(θ) + 1 from both sides)
2
cot (θ) – cot(θ) = 0
(factor cot(θ))
cot(θ)[cot(θ) – 1] = 0
(solve)
cot(θ) = 0 or
cot(θ) – 1 = 0
cot(θ) = 0 or
cot(θ) = 1
π
2
cot(θ) = 0 when cos(θ) = 0 or θ =
or θ =
π
4
or π +
π
4
=
5π
4
or
3π
2
. cot(θ) = 1 when tan(θ) = 1
. None of these values match our
π
restrictions, so the solution is { ,
€
€4
π
2
,
5π
4
,
3π
2
}.
€ €
Ex.€13
sec(θ)
= tan(θ) + cot(θ)
Solution:
€ function
€ € is
€ undefined when the cos(θ) = 0
The secant and tangent
or θ =
π
2
or
3π
2
. The cotangent function is undefined when the
sin(θ) = 0 or θ = 0 or π. Thus, our restrictions are θ ≠ 0,
€
€
€
€
π
2
, π,
3π
2
.
sec(θ) = tan(θ) + cot(θ)
(write in terms of sine and cosine)
1
sin(θ)
cos(θ)
=
+
cos(θ)
cos(θ)
sin(θ)
1
sin(θ)
• cos(θ)sin(θ) =
•
cos(θ)
cos(θ)
(multiply by cos(θ)sin(θ))
€
cos(θ)sin(θ) +
sin(θ) = sin2(θ) + cos2(θ)
€sin(θ) =€1
θ=
π
2
€
cos(θ)
•
sin(θ)
cos(θ)sin(θ)
(sin2(θ) + cos2(θ) = 1)
€
€
But, our restrictions say that θ ≠
π
2
, so we have to reject our answer.
Thus, this equation has no solution.
€
€