Slide 6-1 Equations 6.1 Solving Trigonometric Equations 6.2 More on Trigonometric Equations 6.3...

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Equations

6.1 Solving Trigonometric Equations6.2 More on Trigonometric Equations

6.3 Trigonometric Equations Involving Multiples Angles6.4 Parametric Equations and Further Graphing

Chapter 6

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Decide whether the equation is linear or quadratic in form, so you can determine the solution method.

If only one trigonometric function is present, first solve the equation for that function.

If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve.

6.1 Solving Trigonometric Equations

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If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval.

Try using identities to change the form of the equation. If may be helpful to square both sides of the equation first. If this is done, check for extraneous solutions.

4

Example: Linear Method

Solve 2 cos2 x 1 = 0Solution: First, solve for cos x on the unit circle.

2

2

2

2cos 1 0

2cos 1

1cos

2

1cos

2

2cos

2

x

x

x

x

x

45 ,135 ,225 ,315

3 5 7 , , ,

4 4 4 4

x

or

5

Example: Factoring

Solve 2 cos x + sec x = 0Solution

2

12cos 0

cos1

2cos 1 0cos

10

cos1

0cos

xx

xx

x

x

2

2

2

2cos 1 0

2cos 1

1cos

2

1cos

2

x

x

x

x

1cos

2x

Since neither factor of the equation can equal zero, the equation has no solution.

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Example: Factoring

Solve 2 sin2 x + 3sin x + 1 = 0(2 Sin x + 1)(Sin x +1) = 0

Set each factor equal to 0.

a) 2 Sin x + 1 = 0

b) Sin x + 1 = 0

continued

1

2sin 1 0

2sin 1

1sin

21

sin2

7 11,

6 6

x

x

x

x

x

7

Example: Factoring continued

1

sin 1 0

sin 1

sin ( 1)

3

2

x

x

x

x

7 11 32 , 2 , 2

6 6 2x k k k

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Example: Squaring

Solve cos x + 1 = sin x [0, 2]

2 2

2 2

2

cos 1 sin

cos 2cos 1 sin

cos 2cos 1 1 cos

2cos 2cos 0

2cos (cos 1) 0

2cos 0 and cos 1 0

cos 0 cos 1

3,

2 2

x x

x x x

x x x

x x

x x

x x

x x

x x

Check the solutions in the original equation. The only solutions are /2 and .

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a) over the interval [0, 2), and

b) give all solutions

Solution:

Write the interval as the inequality

Solve 2sin 12

x

0 2 . x

6.2 More on Trigonometric Equations

10

Example: Using a Half-Angle Identity continued

The corresponding interval for x/2 is Solve

Sine values that corresponds to 1/2 are

2sin 12

x

0 .2

x

2sin 12

1sin

2 2

x

x

5 and

6 6

5 or

2 6 2 65

or 3 3

x x

x x

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Example: Using a Half-Angle Identity continued

b) Sine function with a period of 4, all solutions are given by the expressions

where n is any integer.

54 and 4

3 3

n n

12

Example: Double Angle

Solve cos 2x = cos x over the interval [0, 2).

First, change cos 2x to a trigonometric function of x. Use the identity 2cos 2 2cos 1. x x

2

cos 2

2cos

cos

cos1

x

x

x

x

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Example: Double Angle continued

Over the interval

2

2

cos

cos

2cos cos 1 0

(2cos 1)(cos 1) 0

2cos 1 0 or cos 1 0

1cos or

cos 2

cos 12

2cos 1

x

x

x

x

x

x

x x

x x

x x

2 4 or or 0.

3 3

x x x

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Example: Multiple-Angle Identity

Solve over the interval [0, 360).4sin cos 3

2

2

s

si

in

n co

c

s s

os

in 2

4sin cos 3

2( ) 3

2 3

3sin

in 2

22

s

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List all solutions in the interval.

The final two solutions were found by adding 360 to 60 and 120, respectively, giving the solution set

2 60 ,120 ,420 ,480

30 ,60 ,210 ,240

or

30 ,60 ,210 ,240 .

6.3 Trigonometric Equations Involving Multiples Angles

16

Example: Multiple Angle

Solve tan 3x + sec 3x = 2 over the interval [0, 2).

Tangent and secant are related so use the identity 2 21 tan sec .

22

2 2

tan 3 sec3 2

tan 3 2 sec3

4 4stan 3

s

ec3 sec 3

4 4sec3 sec3 3ec 1

x x

x x

xx

x xx

x

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Example: Multiple Angle continued

2 24 4sec3 sec 3

4sec3 5

5sec3

41 5

sec 3

cos3 44

cos3

1

5

x xx

x

x

x

x

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Example: Multiple Angle continued

Use a calculator and the fact that cosine is positive in quadrants I and IV,

Since both sides of the equation were squared, each proposed solution must be checked. The solution set is {.2145, 2.3089, 4.4033}.

3 .6435,5.6397,6.9267,11.9229,13.2099,18.2061

.2145,1.8799,2.3089,3.9743,4.4033,6.0687.

x

x

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Solving for x in Terms of y Using Inverse Function

Example: y = 3 cos 2x for x.

Solution: We want 2x alone on one side of the equation so we

can solve for 2x, and then for x.

3cos 2

cos 23

2 arccos3

1arccos

2 3

y x

yx

yx

yx

20

Solving an Equation Involving an Inverse Trigonometric Function

Example: Solve 2 arcsin

Solution: First solve for arcsin x, and then for x.

The solution set is {1}.

.x

2arcsin

arcsin2

sin2

1

x

x

x

x

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Solving an Equation Involving Inverse Trigonometric Functions

Example: Solve

Solution: Let Then sin and for u in

quadrant I, the equation becomes

1 1 1cos sin .

2x

1 1n .si

2u

1

2u

1cos

cos .

ux

u x

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Solving an Equation Involving Inverse Trigonometric Functions continued

Sketch a triangle and label it using the facts that u is in quadrant I and

Since x = cos u, x = and the solution set is { }.

1sin .

2u

3

2

3,

2

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Solving an Inverse Trigonometric Equation Using an Identity

Example: Solve

Solution: Isolate one inverse function on one side of the equation.

(1)

arcsin arccos .6

x x

arcsin arccos6

arcsin arccos6

sin arccos6

x x

x x

x x

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Solving an Inverse Trigonometric Equation Using an Identity continued

Let u = arccos x, so 0 u by definition.

(2)

Substitute this result into equation (2) to get

(3)

sin6

sin sin cos cos sin6 6 6

xu

u u u

sin cos cos sin .6 6

u u x

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From equation (1) and by the definition of the arcsine function,

Since we must have

Thus x > 0. From this triangle we find that

arccos2 6 22

arccos .3 3

x

x

0 arccos ,x 0 arccos .3

x

2sin 1 .u x

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Now substituting into equation (3) using

The solution set is { }.

2 1sin 1 , sin ,

6 2

3cos , and cos .

6 2

u x

u x

2

2

sin cos cos sin6 6

3 11

2 2

1 3 2

u u x

x x x

x x x

2

2 2

2 2

2

3 1

3 1

3 3

3 4

3

4

3

2

x x

x x

x x

x

x

x

3

2

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A plane curve is a set of points (x, y) such that x = f(t), y = g(t), and f and g are both defined on an interval I. The equations x = f(t) and y = g(t) are parametric equations with parameter t.

6.4 Parametric Equations and Further Graphing

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Graphing a Plane Curve Defined Parametrically

Example: Let x = t2 and y = 2t + 3, for t in [3,3]. Graph the set of ordered pairs (x, y).

Solution: Make a table of

corresponding values of

t, x, and y over the domain of t.

993

742

511

300

111142393yxt

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Graphing a Plane Curve Defined Parametrically continued

Plotting the points shows a graph of a portion of a parabola with horizontal axis y = 3. The arrowheads indicate the direction the curve traces as t increases.

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Finding an Equivalent Rectangular Equation

Example: Find a rectangular equation for the plane curve of the previous example

defined as follows.

x = t2 , y = 2t + 3, for t in [3, 3]

Solution: Solve either equation for t. 2 3

2 3

3

2

y

y

t

t

ty

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Finding an Equivalent Rectangular Equation continued

Now substitute this result into the first equation to get

This is the equation of a horizontal parabola opening to the right. Because t is in [3, 3], x is in [0, 9] and y is in [3, 9]. This rectangular equation must be given with its restricted domain as 4x = (y 3)2 , for x in [0, 9].

22

22 33 or 4 3 .

2 4yt

yyx x

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Graphing a Plane Curve Defined Parametrically

Example: Graph the plane curve defined by x = 2 sin t, y = 3cos t, for t in [0,2 ].

Solution: Use the fact that sin2 t + cos2t = 1. Square both sides of each equation; solve one for sin2 t, the other for cos2t.

2

22

2

2sin

4si

4s

n

in

x t

x

tx

t

2

22

2

3cos

9co

9c

s

os

y t

y

ty

t

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Graphing a Plane Curve Defined Parametrically continued

Now add corresponding sides of the two equations.

This is the equation of an ellipse.

2 2

2 2

2 2

sin c

4

9

19

4os

x y

t tx y

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Finding Alternative Parametric Equation Forms

Give two parametric representations for the equation of the parabola y = (x + 5)2 +3.

Solution: The simplest choice is to let

x = t, y = (t + 5)2 + 3 for t in (, ) Another choice, which leads to a simpler

equation for y, is x = t + 5, y = t2 + 3 for t in (, ).

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Application

A small rocket is launched from a table that is 3.36 ft above the ground. Its initial velocity is 64 ft per sec, and it is launched at an angle of 30° with respect to the ground. Find the rectangular equation that models its path. What type of path does the rocket follow?

Solution: The path of the rocket is defined by the parametric equations

x = (64 cos 30°)t and y = (64 sin 30°)t 16t2 + 3.36

Or equivalently,

232 3 and 16 32 3.36.x t y t t

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Application continued

From we obtain

Substituting into the other parametric equations for t yields

Simplifying, we find that the rectangular equation is

Because the equation defines a parabola, the rocket follows a parabolic path.

32 3 ,x t .32 3

xt

2

16 32 3.36.32 3 32 3

x xy

21 33.36.

192 3y x x

Slide 6-1 Equations 6.1 Solving Trigonometric Equations 6.2 More on Trigonometric Equations 6.3...

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