Transcript 5.3 Solving Trig Equations

Digital Lesson
Solving Trigonometric
Equations
sin x =
1
2
is a trigonometric equation.
x = 6π is one of infinitely many solutions of y = sin x.
y
-19π
6
-3π
-11π
6
-7π
6
π
6
1
5π
6
π
-2π -π
13π
6
17π
6
2π 3π
25π
6
4π
1
y
=
x
2
-1
All the solutions for x can be expressed in the form of
a general solution.
x = 6π + 2kπ and x = 5 6π + 2kπ (k = 0, ±1, ± 2, ± 3,  ).
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2
Find the general solution for the equation sec  = 2.
1
From cos  = 1 , it follows that cos  = .
sec 
2
1
π
1
y
cos( 3 + 2kπ) =
All values of  for which cos  =
2
P
2
are solutions of the equation.
Two solutions are  = ± π .
1
3
x
1
All angles that are coterminal
2
with ± π are also solutions and
3
can be expressed by adding integer
Q
-π + 2kπ) = 1
cos(
multiples of 2π.
3
2
The general solution can be written as  = ± π + 2kπ .
3
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3
Example: Solve tan x = 1.
The graph of y = 1 intersects the graph of y = tan x infinitely
y
many times.
- π – 2π
4
-π–π
4
π
π + π π + 2π
π + 3π
4
4
4
4
y=1
-π
π
x
2π
3π
y = tan(x)
x = -3π x = -π
2
2
x = π x = 3π x = 5π
2
2
2
Points of intersection are at x = π and every multiple of π added or
4
π
subtracted from 4 .
General solution: x = π + kπ for k any integer.
4
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4
Example:
Solve the equation 3sin x + 2 = sin x for  π ≤ x ≤ π .
2
2
y
3sin x + 2 = sin x
3sin x  sin x + 2 = 0
2sin x + 2 = 0
Collect like terms.
sin x =  2
2
1
-π
1 4
x
y=- 2
2
x = 4π is the only solution in the interval  2π ≤ x ≤ 2π .
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5
Example: To find all solutions of cos4(2x) = 9 .
16
Take the fourth root of both sides to obtain:
cos(2x)= ± 3
y
2
From the unit circle, the
solutions for 2 are
2 = ± π + kπ,
6
k any integer.
π
π
1
π
6
x
-π
6
x=- 3
2
x=
3
2
π + k ( π ), for k any integer.
Answer:  = ± 12
2
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6
Find all solutions of the trigonometric equation:
tan2  + tan  = 0.
tan2  + tan  = 0
tan  (tan  +1) = 0
Original equation
Factor.
Therefore, tan  = 0 or tan  = -1.
The solutions for tan  = 0 are the values  = kπ,
for k any integer.
The solutions for tan  = 1 are  = - 4π + kπ,
for k any integer.
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7
The trigonometric equation 2 sin2  + 3 sin  + 1 = 0 is
quadratic in form.
2 sin2  + 3 sin  + 1 = 0 implies that
(2 sin  + 1)(sin  + 1) = 0.
Therefore, 2 sin  + 1 = 0 or sin  + 1 = 0.
It follows that sin  = - 1 or sin  = -1.
2
Solutions:
 = - π + 2kπ and  = 7π + 2kπ, from sin  = - 1
6
6
2
 = -π + 2kπ, from sin  = -1
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8
Example: Solve 8 sin  = 3 cos2  with  in the
interval [0, 2π].
Rewrite the equation in terms of only one trigonometric function.
8 sin  = 3(1 sin2  )
Use the Pythagorean Identity.
3 sin2  + 8 sin   3 = 0.
A “quadratic” equation with sin x
as the variable
(3 sin   1)(sin  + 3) = 0 Factor.
Therefore, 3 sin   1 = 0 or sin  + 3 = 0
Solutions: sin  = 1 or sin  = -3
3
 = sin1( 1 ) = 0.3398 and  = π  sin1( 1 ) = 2.8107.
3
3
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s
9
Solve: 5cos2  + cos  – 3 = 0 for 0 ≤  ≤ π.
The equation is quadratic. Let y = cos  and solve 5y2 + y  3 = 0.
y = (-1 ± 61 ) = 0.6810249 or -0.8810249
10
Therefore, cos  = 0.6810249 or –0.8810249.
Use the calculator to find values of  in 0 ≤  ≤ π.
This is the range of the inverse cosine function.
The solutions are:
 = cos 1(0.6810249 ) = 0.8216349 and
 = cos 1(0.8810249) = 2.6488206
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10
Example: Find the intersection points of the graphs of
y
y = sin  and y = cos .
π + kπ
4
The two solutions
for  between 0 and
2π are π and 5π .
4
4
5 -π
4
π + kπ
4
1
-π
4
x
1
The graphs of y = sin  and y = cos  intersect at
points where sin  = cos .
This is true only for 45-45-90 triangles.
The general solution is  = 4π + kπ, for k any integer.
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11
HELPFUL HINTS FOR SOLVING
TRIGONOMETRIC EQUATIONS
•Try to get equations in terms of one trig function by using
identities.
•Be on the look-out for ways to substitute using identities
•Try to get trig functions of the same angle. If one term is
cos2 and another is cos for example, use the double angle
formula to express first term in terms of just  instead of 2
•Get one side equals zero and factor out any common trig
functions
•See if equation is quadratic in form and will factor. (replace
the trig function with x to see how it factors if that helps)
•If the angle you are solving for is a multiple of , don't forget
to add 2 to your answer for each multiple of  since  will
still be less than 2 when solved for.
There are some equations that can't be solved by hand
and we must use a some kind of technology.
Use a graphing utility to solve the equation. Express any solutions
rounded to two decimal places.
22x 17 sin x  3
Graph this side as y1
in your calculator
Graph this side as y2
in your calculator
You want to know where they are equal. That would be
where their graphs intersect. You can use the trace
feature or the intersect feature to find this (or these)
points (there could be more than one point of
intersection).
22x 17 sin x  3
This
This was
is offgraphed
a little due
on
the
to the
computer
fact we with
check: 22 .53  17sin .53  3.066  3
graphcalc,
approximated.
a freeIf you
graphing
carried it utility
to more
you
can
decimal
download
placesatyou'd
www.graphcalc.com
have more accuracy.
After seeing the initial
graph, lets change the
window to get a better
view of the intersection
point and then we'll do
a trace.
Rounded to 2 decimal places, the
point of intersection is x = 0.53
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating some of this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au
Text Example
Solve the equation:
2 cos2 x + cos x  1  0,
0  x < 2.
Solution
The given equation is in quadratic form 2t2 + t  1  0 with t 
cos x. Let us attempt to solve the equation using factoring.
2 cos2 x + cos x  1  0
This is the given equation.
(2 cos x  1)(cos x + 1)  0
2 cos x  1 0
or
Factor. Notice that 2t2 + t – 1 factors as (2t – 1)(2t + 1).
cos x + 1  0
2 cos x  1 cos x  1
Set each factor equal to 0.
Solve for cos x.
cos x  1/2
x   x  2   x  
The solutions in the interval [0, 2) are /3, , and 5/3.
Example
• Solve the following equation:
7 cos + 9  2 cos
Solution:
7 cos + 9  2 cos
9 cos  9
cos  1
   ,3 ,5
   + 2n
Example
• Solve the equation on the interval [0,2)

3
tan 
2
3
Solution:

3
tan 
2
3


7
 and
2 6
6

7
  and
3
3
Example
• Solve the equation on the interval [0,2)
cos x + 2 cos x  3  0
2
Solution:
cos 2 x + 2 cos x  3  0
(cos x + 3)(cos x  1)  0
cos x + 3  0 cos x  1  0
cos x  3 cos x  1
no solution
x0
x0
Example
• Solve the equation on the interval [0,2)
sin 2x  sin x
Solution:
sin 2 x  sin x
2 sin x cos x  sin x
2 cos x  1
1
cos x 
2

5
x ,
3
3