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P-BLTZMC05_585-642-hr 21-11-2008 12:54 Page 626
626 Chapter 5 Analytic Trigonometry
Section 5.5 Trigonometric Equations
Objectives
Find all solutions of a xponential functions display the manic
trigonometric equation. Eenergies of uncontrolled growth. By
contrast, trigonometric functions repeat
Solve equations with their behavior. Do they embody in their
multiple angles. regularity some basic rhythm of the
Solve trigonometric equations universe? The cycles of periodic phenomena
quadratic in form. provide events that we can comfortably
Use factoring to separate count on.When will the moon look just as
different functions in it does at this moment? When can I count
trigonometric equations. on 13.5 hours of daylight? When will my
Use identities to solve breathing be exactly as it is right now?
trigonometric equations. Models with trigonometric functions
embrace the periodic rhythms of our
Use a calculator to solve world.Equations containing trigonometric
trigonometric equations. functions are used to answer questions
about these models.
Find all solutions of a Trigonometric Equations and Their Solutions
trigonometric equation. A trigonometric equation is an equation that contains a trigonometric expression
with a variable, such as sin x. We have seen that some trigonometric equations are
2 2
identities, such as sin x + cos x = 1. These equations are true for every value of
the variable for which the expressions are defined. In this section, we consider
trigonometric equations that are true for only some values of the variable. The
values that satisfy such an equation are its solutions. (There are trigonometric
equations that have no solution.)
An example of a trigonometric equation is
sin x = 1.
2
p p 1
A solution of this equation is 6 because sin 6 = 2. By contrast, p is not a solution
because sin p = 0 Z 1.
2
Is p the only solution of sin x = 1? The answer is no. Because of the periodic
6 2 1
nature of the sine function,there are infinitely many values of x for which sin x = 2.
p 3p 7p
Figure 5.7 shows five of the solutions, including 6, for - 2 … x … 2 . Notice that
the x-coordinates of the points where the graph of y = sin x intersects the line
y = 1 are the solutions of the equation sin x = 1.
2 2
y y = sin x
y = 1 1
2
Figure 5.7 The equation sin x = 1 x
2 −' k l m x
has five solutions when x is restricted to
the interval c- 3p, 7pd. −1
2 2 solution solution solution solution solution
How do we represent all solutions of sin x = 1? Because the period of the sine
2
function is 2p, first find all solutions in 30, 2p2. The solutions are
p p 5p
x= and x=p-= .
6 6 6
The sine is positive in quadrants I and II.
P-BLTZMC05_585-642-hr 21-11-2008 12:54 Page 627
Section 5.5 Trigonometric Equations 627
Any multiple of 2p can be added to these values and the sine is still 1. Thus, all
solutions of sin x = 1 are given by 2
2
x = p + 2np or x = 5p + 2np,
6 6
where n is any integer. By choosing any two integers, such as n = 0 and n = 1, we
can find some solutions of sin x = 1. Thus, four of the solutions are determined as
2
follows:
Let n = 0. Let n = 1.
p 5p p 5p
x=6+2 0p x= 6+2 0p x=6+2 1p x=6+2 1p
p 5p p 5p
=6 =6 =6+2p x=6+2p
p 12p 13p 5p 12p 17p
==++= = .
6 6 6 6 6 6
These four solutions are shown among the five solutions in Figure 5.7.
Equations Involving a Single Trigonometric Function
To solve an equation containing a single trigonometric function:
• Isolate the function on one side of the equation.
• Solve for the variable.
EXAMPLE 1 Finding All Solutions of a Trigonometric Equation
Solve the equation: 3 sin x - 2 = 5 sin x - 1.
Solution The equation contains a single trigonometric function,sin x.
Step 1 Isolate the function on one side of the equation. We can solve for sin x
by collecting terms with sin x on the left side and constant terms on the right side.
3 sin x - 2 = 5 sin x - 1 This is the given equation.
3 sin x - 5 sin x - 2 = 5 sin x - 5 sin x - 1 Subtract 5 sin x from both sides.
-2 sin x - 2 =-1 Simplify.
-2 sin x = 1 Add 2 to both sides.
1
sin x =- Divide both sides by -2 and solve
2
for sin x.
1
Step 2 Solve for the variable. We must solve for x in sin x =- . Because
p 1 1 2
sin = ,the solutions of sin x =- in 30, 2p2 are
6 2 2
p 6p p 7p p 12p p 11p
x=p+=+= x=2p-=- = .
6 6 6 6 6 6 6 6
The sine is negative The sine is negative
in quadrant III. in quadrant IV.
Because the period of the sine function is 2p, the solutions of the equation are
given by
x = 7p + 2np and x = 11p + 2np,
6 6
where n is any integer.
P-BLTZMC05_585-642-hr 21-11-2008 12:54 Page 628
628 Chapter 5 Analytic Trigonometry Solve the equation: 5 sin x = 3 sin x + 23.
Check Point 1
Now we will concentrate on finding solutions of trigonometric equations for
0 … x 6 2p.You can use a graphing utility to check the solutions of these equations.
Graph the left side and graph the right side.The solutions are the x-coordinates of the
points where the graphs intersect.
Solve equations with multiple Equations Involving Multiple Angles
angles. Here are examples of two equations that include multiple angles:
tan 3x =1 x 3
sin 2 =.
2
The angle is a The angle is a
multiple of 3. 1
multiple of .
2
We will solve each equation for 0 … x 6 2p. The period of the function plays an
important role in ensuring that we do not leave out any solutions.
EXAMPLE 2 Solving an Equation with a Multiple Angle
Solve the equation: tan 3x = 1, 0 … x 6 2p.
Solution The period of the tangent function is p. In the interval 30, p2, the only
value for which the tangent function is 1 is p.This means that 3x = p. Because the
4 4
period is p, all the solutions to tan 3x = 1 are given by
Technology 3 x = p + np. n is any integer.
Graphic Connections 4
x = p + np Divide both sides by 3 and solve for x.
Shown below are the graphs of 12 3
y = tan 3x In the interval 30, 2p2, we obtain the solutions of tan 3x = 1 as follows:
and Let n = 0. Let n = 1. Let n = 2.
y = 1 p 0p p 1p p 2p
x=12+ 3 x=12+ 3 x=12+ 3
in a c0, 2p, pd by 3-3, 3, 14 viewing
2 =p =p+4p=5p =p+8p=9p=3p
rectangle.The solutions of 12 12 12 12 12 12 12 4
tan 3x = 1 Let n = 3. Let n = 4. Let n = 5.
in 30, 2p2 are shown by the
x-coordinates of the six intersection x=p+3p x=p+4p x=p+5p
points. 12 3 12 3 12 3
p 12p 13p p 16p 17p p 20p 21p 7p
=12+ = =12+ = =12+ =.=
12 12 12 12 12 12 4
If you let n = 6, you will obtain x = 25p. This value exceeds 2p. In the interval
12
30, 2p2, the solutions of tan 3x = 1 are p , 5p, 3p, 13p, 17p, and 7p. These
12 12 4 12 12 4
solutions are illustrated by the six intersection points in the technology box.
Check Point 2 Solve the equation: tan 2x = 23, 0 … x 6 2p.
EXAMPLE 3 Solving an Equation with a Multiple Angle
x 23
Solve the equation: sin 2 = 2 , 0 … x 6 2p.
P-BLTZMC05_585-642-hr 21-11-2008 12:54 Page 629
Section 5.5 Trigonometric Equations 629
Solution The period of the sine function is 2p. In the interval 30, 2p2, there are two
23 p
values at which the sine function is 2 . One of these values is 3 .The sine is positive
in quadrant II; thus, the other value is p - p, or 2p. This means that x = p
x 2p 3 3 x 23 2 3
or 2 = 3 .Because the period is 2p, all the solutions of sin 2 = 2 are given by
x p x 2p
= + 2np or = + 2np n is any integer.
2 3 2 3
x = 2p + 4np x = 4p + 4np. Multiply both sides by 2 and solve for x.
3 3
We see that x = 2p + 4npor x = 4p + 4np.If n = 0,we obtain x = 2p from the
3 3 3
first equation and x = 4p from the second equation.If we let n = 1, we are adding
3
4 # 1 # p, or 4p, to 2p and 4p. These values of x exceed 2p. Thus, in the interval
3 3 x 23 2p 4p
30, 2p2, the only solutions of sin 2 = 2 are 3 and 3 .
Check Point 3 Solve the equation: sin x = 1, 0 … x 6 2p.
3 2
Solve trigonometric equations Trigonometric Equations Quadratic in Form
quadratic in form. Some trigonometric equations are in the form of a quadratic equation
2
au + bu + c = 0, where u is a trigonometric function and a Z 0. Here are two
examples of trigonometric equations that are quadratic in form:
2 cos2 x+cos x-1=0 2 sin2 x-3 sin x+1=0.
Technology The form of this equation is The form of this equation is
2 2
2u + u − 1 = 0 with u = cos x. 2u − 3u + 1 = 0 with u = sin x.
Graphic Connections
To solve this kind of equation, try using factoring. If the trigonometric expression does
The graph of not factor,use another method,such as the quadratic formula or the square root property.
2
y = 2 cos x + cos x - 1
is shown in a EXAMPLE 4 Solving a Trigonometric Equation Quadratic in Form
p Solve the equation: 2 x + cos x - 1 = 0, 0 … x 6 2p.
c0, 2p, d by 3-3, 3, 14 2 cos
2 2 + u - 1 = 0
Solution The given equation is in quadratic form 2u with
viewing rectangle. The x-intercepts, u = cos x. Let us attempt to solve the equation by factoring.
p, p, and 5p, 2 cos2 x + cos x - 1 = 0 This is the given equation.
3 3 12 cos x - 121cos x + 12 = 0 Factor: Notice that 2u2 + u - 1
verify the three solutions of factors as 12u - 121u + 12.
2
2 cos x + cos x - 1 = 0 2 cos x - 1 = 0 or cos x + 1 = 0 Set each factor equal to 0.
in 30, 2p2. 2 cos x = 1 cos x =-1 Solve for cos x.
cos x = 1
y = 2 cos2 x + cos x − 1 p 2 p 5p
x=3 x=2p-3=3 x=p Solve each equation for x,
O … x 6 2p.
The cosine is positive
x = p x = p x = 5p in quadrants I and IV.
3 3
The solutions in the interval 30, 2p,2 are p, p, and 5p.
3 3
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