Download Pre-Class Problems 18

Pre-Class Problems 18 for Friday, March 31
These are the type of problems that you will be working on in class. These
problems are from Lesson 10.
You can go to the solution for each problem by clicking on the problem letter.
Objective of the following problems: To find all the solutions to a trigonometric
equation.
1.
Find all the exact solutions for the following equations.
a.
sin   
3 cot   1  0
c.
e.
3
2
2 cos  
3  0
g. cos   0
b.
2 cos  
d.
3 csc x  4  2
f.
9 tan   2  7
h.
5 sin   8  3
2  0
i.
6 sec   5  1
j.
sin x  0
k.
tan    4
l.
8 sec   15   3
Additional problems available in the textbook: Page 618 … 9 – 18, 57 – 60.
Examples 1 and 11 starting on page 608.
Solutions:
1a.
sin   
3
2
Back to Problem 1.
NOTE: Sine is negative in the third and fourth quadrants. If we find one
angle solution in the third quadrant, then all the other angle solutions in the
third quadrant are coterminal to this one. The same is true for all the
solutions in the fourth quadrant.
To find the reference angle  ' :
sin   
3
 sin  ' 
2
3
3

  '  sin  1

2
2
3
NOTE: The one solution in the third quadrant that is in the interval [ 0 , 2  )

3

4



is given by      '   
.
3
3
3
3
All solutions in III:  
4
 2 n  , where n is an integer
3
RECALL: The set of integers is the set { .... ,  3 ,  2 ,  1, 0 , 1, 2 , 3 , .... } .
NOTE: The one solution in the fourth quadrant that is in the interval [ 0 , 2  )

6

5



is given by   2    '  2  
. The one solution
3
3
3
3
in the fourth quadrant that is in the interval (  2  , 0 ] is given by
  '  

.
3
All solutions in IV:  
5
 2 n
3
or   

 2 n  , where n is an
3
integer
Answer:  
4
5
 2 n ,  
 2 n  , where n is an integer
3
3
OR  
1b.
2 cos  
4

 2 n ,   
 2 n  , where n is an integer
3
3
2  0
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate cos  on one
side of the equation by itself.
2 cos  
2  0  2 cos  
2  cos  
2
2
NOTE: Cosine is positive in the first and fourth quadrants. If we find one
angle solution in the first quadrant, then all the other angle solutions in the
first quadrant are coterminal to this one. The same is true for all the solutions
in the fourth quadrant.
To find the reference angle  ' :
cos  
2
 cos  ' 
2
2
2

  '  cos  1

2
2
4
NOTE: The one solution in the first quadrant that is in the interval [ 0 , 2  )
is given by    ' 
All solutions in I:  

.
4

 2 n  , where n is an integer
4
NOTE: The one solution in the fourth quadrant that is in the interval [ 0 , 2  )

8

7



is given by   2    '  2  
. The one solution
4
4
4
4
in the fourth quadrant that is in the interval (  2  , 0 ] is given by
  '  

.
4
All solutions in IV:  
7

 2 n  or   
 2 n  , where n is an
4
4
integer
Answer:  

7
 2 n ,  
 2 n  , where n is an integer
4
4
OR  
1c.


 2 n ,   
 2 n  , where n is an integer
4
4
3 cot   1  0
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate cot  on one
side of the equation by itself.
3 cot   1  0 
3 cot    1  cot   
1
3
NOTE: Now, we will take the reciprocal of both sides in order to obtain
tan  on the left side of the equation.
cot   
1
 tan   
3
3
NOTE: Tangent is negative in the second and fourth quadrants. If we find
one angle solution in the second quadrant, then all the other angle solutions in
the second quadrant are coterminal to this one. The same is true for all the
solutions in the fourth quadrant.
To find the reference angle  ' :
tan   
3  tan  ' 
3   '  tan  1 3 

3
NOTE:
The one solution in the second quadrant that is in the interval

3

2



[ 0 , 2  ) is given by      '   
.
3
3
3
3
All solutions in II:  
2
 2 n  , where n is an integer
3
NOTE: The one solution in the fourth quadrant that is in the interval [ 0 , 2  )

6

5



is given by   2    '  2  
. The one solution
3
3
3
3
in the fourth quadrant that is in the interval (  2  , 0 ] is given by
   '  

.
3
All solutions in IV:  
5

 2 n  or   
 2 n  , where n is an
3
3
integer
Answer:  
2
5
 2 n ,  
 2 n  , where n is an integer
3
3
OR  
2

 2 n ,   
 2 n  , where n is an integer
3
3
2
5
 2 n ,  
 2 n  , where n is an
3
3
2
 n  , where n is an integer.
integer, may also be written as  
3
NOTE: The answer  
1d.
3 csc x  4  2
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate csc x on one
side of the equation by itself.
3 csc x  4  2  3 csc x  6  csc x  2
NOTE: Now, we will take the reciprocal of both sides in order to obtain
sin x on the left side of the equation.
csc x  2  sin x 
1
2
NOTE: Sine is positive in the first and second quadrants. If we find one
angle solution in the first quadrant, then all the other angle solutions in the
first quadrant are coterminal to this one. The same is true for all the solutions
in the second quadrant.
To find the reference angle x ' :
sin x 
1
1
1

 sin x ' 
 x '  sin  1 
2
2
2
6
NOTE: The one solution in the first quadrant that is in the interval [ 0 , 2  )
is given by x  x ' 

.
6
All solutions in I: x 
NOTE:

 2 n  , where n is an integer
6
The one solution in the second quadrant that is in the interval

6

5



[ 0 , 2  ) is given by x    x '   
.
6
6
6
6
All solutions in II: x 
Answer: x 
1e.
2 cos  
5
 2 n  , where n is an integer
6

5
 2 n , x 
 2 n  , where n is an integer
6
6
3  0
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate cos  on one
side of the equation by itself.
2 cos  
3  0  2 cos   
3  cos   
3
2
NOTE: Cosine is negative in the second and third quadrants. If we find one
angle solution in the second quadrant, then all the other angle solutions in the
second quadrant are coterminal to this one. The same is true for all the
solutions in the third quadrant.
To find the reference angle  ' :
cos   
3
 cos  ' 
2
3
3

  '  cos  1

2
2
6
NOTE:
The one solution in the second quadrant that is in the interval

6

5



[ 0 , 2  ) is given by      '   
.
6
6
6
6
All solutions in II:  
5
 2 n  , where n is an integer
6
NOTE: The one solution in the third quadrant that is in the interval [ 0 , 2  )

6

7



is given by      '   
.
6
6
6
6
All solutions in III:  
Answer:  
1f.
7
 2 n  , where n is an integer
6
5
7
 2 n ,  
 2 n  , where n is an integer
6
6
9 tan   2  7
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate tan  on one
side of the equation by itself.
9 tan   2  7  9 tan   9  tan   1
NOTE: Tangent is positive in the first and third quadrants. If we find one
angle solution in the first quadrant, then all the other angle solutions in the
first quadrant are coterminal to this one. The same is true for all the solutions
in the third quadrant.
To find the reference angle  ' :
tan   1  tan  '  1   '  tan  1 1 

4
NOTE: The one solution in the first quadrant that is in the interval [ 0 , 2  )
is given by    ' 
All solutions in I:  

.
4

 2 n  , where n is an integer
4
NOTE: The one solution in the third quadrant that is in the interval [ 0 , 2  )

4

5



is given by      '   
.
4
4
4
4
All solutions in III:  
Answer:  
5
 2 n  , where n is an integer
4

5
 2 n ,  
 2 n  , where n is an integer
4
4
NOTE: This answer may also be written as  

 n  , where n is an
4
integer.
1g.
cos   0
Back to Problem 1.
NOTE: The angle solutions for this equation must lie on one of the
coordinate axes since zero is not positive nor negative.
NOTE: Using Unit Circle Trigonometry, we know that cos
the one solution, which lies on the positive y-axis, is  
All solutions on the positive y-axis:  

.
2

 0 . Thus,
2

 2 n  , where n is an integer
2
3
 0 . Thus,
2
3
the one solution, which lies on the negative y-axis, is  
. Also, using
2
NOTE: Using Unit Circle Trigonometry, we know that cos
 
Unit Circle Trigonometry, we know that cos     0 .
 2
solution, which lies on the negative y-axis, is   
All solutions on the negative y-axis:  
Thus, the one

.
2
3

 2 n  or   
 2 n ,
2
2
where n is an integer
Answer:  

3
 2 n ,  
 2 n  , where n is an integer
2
2
OR  


 2 n ,   
 2 n  , where n is an integer
2
2

3
 2 n ,  
 2 n  , where n is an
2
2

 n  , where n is an integer.
integer, may also be written as  
2
NOTE:
1h.
The answer  
5 sin   8  3
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate sin  on one
side of the equation by itself.
5 sin   8  3  5 sin    5  sin    1
NOTE: The angle solutions for this equation must lie on one of the
coordinate axes since  1 is the minimum negative number in the range of the
sine function.
3
  1 . Thus,
2
3


the one solution, which lies on the negative y-axis, is
. Also, using
2
 
Unit Circle Trigonometry, we know that sin      1 . Thus, the one
 2

solution, which lies on the negative y-axis, is    .
2
NOTE: Using Unit Circle Trigonometry, we know that sin
All solutions on the negative y-axis:  
3

 2 n  or   
 2 n ,
2
2
where n is an integer
Answer:  
1i.
3

 2 n  or   
 2 n  , where n is an integer
2
2
6 sec   5  1
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate sec  on one
side of the equation by itself.
6 sec   5  1  6 sec   6  sec   1
NOTE: Now, we will take the reciprocal of both sides in order to obtain
cos  on the left side of the equation.
sec   1  cos   1
NOTE: The angle solutions for this equation must lie on one of the
coordinate axes since 1 is the maximum positive number in the range of the
cosine function.
NOTE: Using Unit Circle Trigonometry, we know that cos 0  1 . Thus, the
one solution, which lies on the positive x-axis, is   0 .
All solutions on the positive x-axis:   0  2 n   2 n  , where n is an
integer
Answer:   2 n  , where n is an integer
1j.
sin x  0
Back to Problem 1.
NOTE: The angle solutions for this equation must lie on one of the
coordinate axes since zero is not positive nor negative.
NOTE: Using Unit Circle Trigonometry, we know that sin 0  0 . Thus, the
one solution, which lies on the positive x-axis, is x  0 .
All solutions on the positive x-axis: x  0  2 n   2 n  , where n is an
integer
NOTE: Using Unit Circle Trigonometry, we know that sin   0 . Thus, the
one solution, which lies on the negative x-axis, is x   .
All solutions on the negative x-axis: x    2 n   ( 2 n  1 )  , where n
is an integer
Answer: x  2 n  , x  ( 2 n  1 )  , where n is an integer
NOTE: This answer may also be written as x  n  , where n is an integer.
1k.
tan    4
Back to Problem 1.
NOTE: Tangent is negative in the second and fourth quadrants. If we find
one angle solution in the second quadrant, then all the other angle solutions in
the second quadrant are coterminal to this one. The same is true for all the
solutions in the fourth quadrant.
To find the reference angle  ' :
tan    4  tan  '  4   '  tan  1 4
NOTE: The one solution in the second quadrant that is in the interval
[ 0 , 2  ) is given by      '    tan  1 4 .
All solutions in II:
where n is an integer
    tan  1 4  2 n    tan  1 4  ( 2 n  1)  ,
NOTE: The one solution in the fourth quadrant that is in the interval [ 0 , 2  )
1
is given by   2    '  2   tan 4 . The one solution in the fourth
1
quadrant that is in the interval (  2  , 0 ] is given by     '   tan 4 .
1
1
All solutions in IV:   2   tan 4  2 n    tan 4  2 ( n  1) 
1
or    tan 4  2 n  , where n is an integer
Answer:
   tan  1 4  ( 2 n  1)  ,    tan  1 4  2 ( n  1)  ,
where n is an integer
OR
   tan  1 4  ( 2 n  1)  ,    tan  1 4  2 n  ,
where n is an integer
1l.
8 sec   15   3
Back to Problem 1.
NOTE: In order to solve this equation, we must first isolate sec  on one
side of the equation by itself.
8 sec   15   3  8 sec   12  sec  
12
3
 sec  
8
2
NOTE: Now, we will take the reciprocal of both sides in order to obtain
cos  on the left side of the equation.
sec  
3
2
 cos  
2
3
NOTE: Cosine is positive in the first and fourth quadrants. If we find one
angle solution in the first quadrant, then all the other angle solutions in the
first quadrant are coterminal to this one. The same is true for all the solutions
in the fourth quadrant.
To find the reference angle  ' :
cos  
2
2
2
 cos  ' 
  '  cos  1
3
3
3
NOTE: The one solution in the first quadrant that is in the interval [ 0 , 2  )
1 2



'

cos
is given by
.
3
1
All solutions in I:   cos
2
 2 n  , where n is an integer
3
NOTE:
The one solution in the fourth quadrant that is in the interval
1 2
(  2  , 0 ] is given by     '   cos
.
3
1
All solutions in IV:    cos
1
Answer:   cos
2
 2 n  , where n is an integer
3
2
2
 2 n  ,    cos  1  2 n  , where n is an integer
3
3