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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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