Slide 6-1
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
Slide 6-2
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
Slide 6-3
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.
Slide 6-4
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
Slide 6-5
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.
Slide 6-6
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
Slide 6-7
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
Slide 6-8
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 .
Slide 6-9
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
Slide 6-10
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
Slide 6-11
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
Slide 6-12
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
Slide 6-13
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
Slide 6-14
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
Slide 6-15
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
Slide 6-16
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
Slide 6-17
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
Slide 6-18
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
Slide 6-19
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
Slide 6-20
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
Slide 6-21
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
Slide 6-22
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
Slide 6-23
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
Slide 6-24
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
Slide 6-25
Solving an Inverse Trigonometric Equation Using an Identity continued
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
Slide 6-26
Solving an Inverse Trigonometric Equation Using an Identity continued
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
Slide 6-27
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
Slide 6-28
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
Slide 6-29
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.
Slide 6-30
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
Slide 6-31
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
Slide 6-32
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
Slide 6-33
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
Slide 6-34
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 (, ).
Slide 6-35
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
Slide 6-36
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