Pre-Class Problems 20 for Friday, November 13
These are the type of problems that you will be working on in class. These
problems are from Lesson 10.
Solution to Problems on the Pre-Exam.
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 using square roots or factoring.
1.
Find all the exact solutions for the following equations.
a.
4 cos 2   9  12
b.
c.
sin  ( sec  
d.
3 cos x tan x  cos x
e.
( sin   1 ) ( cos   1 )  0
f.
2 sin 2   3 sin   1  0
g.
6 cos 2   7 cos   5  0
h.
sin 2   24  11sin 
i.
tan 2 x  5 tan x  6  0
j.
3 sec 2   sec   14  0
2)  0
csc 2   2  0
Additional problems available in the textbook: Page 231 … 15 - 27, 29, 30, 33, 34,
47, 63, 64, 69 - 74. Examples 2, 3, and 4 starting on page 225.
Solutions:
1a.
4 cos 2   9  12
4 cos 2   9  12  4 cos 2   3  cos 2  
Back to Problem 1.
3
3
 cos   
4
2
NOTE: There are two equations to solve:
1.
cos  
3
2
and 2.
cos   
3
2
The reference angle  ' for both equations:
cos  ' 
3
3

  '  cos  1

2
2
6
The solutions of the equation cos  
3
:
2
NOTE: Cosine is positive in the first and fourth quadrants.

 2 n  , where n is an integer
6
I:
 
IV:
  

 2 n
6
The solutions of the equation cos   
3
:
2
NOTE: Cosine is negative in the second and third quadrants.
II:
 
5
 2 n  , where n is an integer
6
III:
 
7
 2 n
6
NOTE: The two solutions of  
written as the one solution  

7
 2 n  and  
 2 n  may be
6
6

 n  , where n is an integer.
6
NOTE: The two solutions of   
be written as the one solution   
Answer:

 n  , where n is an integer.
6


5
 2 n ,  
 2 n ,  
 2 n ,
6
6
6
7
 
 2 n  , where n is an integer
6
  
OR   
1b.

5
 2 n  and  
 2 n  may
6
6


 n ,  
 n  , where n is an integer
6
6
csc 2   2  0
Back to Problem 1.
csc 2   2  0  csc 2   2  sin 2  
2
1
1
 sin   
 
2
2
2
NOTE: There are two equations to solve:
1.
sin  
2
2
and 2.
sin   
2
2
The reference angle  ' for both equations:
sin  ' 
2
2

  '  sin  1

2
2
4
The solutions of the equation sin  
2
:
2
NOTE: Sine is positive in the first and second quadrants.
I:
 

 2 n  , where n is an integer
4
II:
 
3
 2 n
4
The solutions of the equation sin   
2
:
2
NOTE: Sine is negative in the third and fourth quadrants.
5
 2 n  , where n is an integer
4
III:
 
IV:
  

 2 n
4
NOTE: The two solutions of  
written as the one solution  

5
 2 n  and  
 2 n  may be
4
4

 n  , where n is an integer.
4
NOTE: The two solutions of   
written as the one solution   
Answer:

 n  , where n is an integer.
4


3
 2 n ,  
 2 n ,  
 2 n ,
4
4
4
5
 
 2 n  , where n is an integer
4
  
OR   
1c.

3
 2 n  and  
 2 n  may be
4
4
sin  ( sec  


 n ,  
 n  , where n is an integer
4
4
2)  0
Back to Problem 1.
NOTE: If a b  0 , then either a  0 , or b  0 , or both are zero.
sin  ( sec  
2 )  0  sin   0 , sec  
2  0
NOTE: There are two equations to solve:
1.
sin   0
and 2.
sec  
2  0
The solutions of the equation sin   0 :
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 and
sin   0 .
Positive x-axis:   0  2 n   2 n  , where n is an integer
Negative x-axis:     2 n   ( 2 n  1 ) 
NOTE: The two solutions of   2 n  and   ( 2 n  1 )  may be written
as the one solution   n  , where n is an integer.
The solutions of the equation sec  
sec  
2  0  sec  
2  0:
2  cos  
1

2
2
2
NOTE: Cosine is positive in the first and fourth quadrants.
To find the reference angle  ' :
cos  
2
2

  '  cos  1

2
2
4

 2 n  , where n is an integer
4
I:
 
IV:
  
Answer:
2
 cos  ' 
2

 2 n
4
  2 n  ,   ( 2 n  1)  ,   
 

 2 n  , where n is an integer
4
OR   n  ,   
an integer

 2 n ,
4


 2 n ,  
 2 n  , where n is
4
4
1d.
3 cos x tan x  cos x
Back to Problem 1.
COMMENT: Most students want to divide both sides of this equation by
cos x . You can’t do this because you might be dividing by zero, which is
undefined. Note that you can subtract cos x from both sides of the equation.
3 cos x tan x  cos x 
3 cos x tan x  cos x  0 
cos x ( 3 tan x  1)  0  cos x  0 ,
3 tan x  1  0
NOTE: There are two equations to solve:
1.
cos x  0
and 2.
3 tan x  1  0
The solutions of the equation cos x  0 :
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
cos
3
 0.
2
Positive y-axis: x 
Negative y-axis: x 

 2 n  , where n is an integer
2
3
 2 n
2

 0 and
2
NOTE: The two solutions of x 

3
 2 n  and x 
 2 n  may be
2
2
written as the one solution x 

 n  , where n is an integer.
2
The solutions of the equation
3 tan x  1  0 :
3 tan x  1  0 
3 tan x  1  tan x 
1
3
NOTE: Tangent is positive in the first and third quadrants.
To find the reference angle x ' :
tan x 
1
 tan x ' 
3
1
1

 x '  tan  1

6
3
3
I:
x 

 2 n  , where n is an integer
6
III:
x 
7
 2 n
6
NOTE: The two solutions of x 
written as the one solution x 
Answer:

7
 2 n  and x 
 2 n  may be
6
6

 n  , where n is an integer.
6

3

 2 n , x 
 2 n , x 
 2 n ,
2
2
6
7
x 
 2 n  , where n is an integer
6
x 
OR x 
1e.


 n , x 
 n  , where n is an integer
2
6
( sin   1 ) ( cos   1 )  0
Back to Problem 1.
( sin   1 ) ( cos   1 )  0  sin   1  0 , cos   1  0
NOTE: There are two equations to solve:
1.
sin   1  0
and 2.
cos   1  0
The solutions of the equation sin   1  0 :
sin   1  0  sin   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
sine function.
NOTE: Using Unit Circle Trigonometry, we know that sin
Positive y-axis:  

 1.
2

 2 n  , where n is an integer
2
The solutions of the equation cos   1  0 :
cos   1  0  cos    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
cosine function.
NOTE: Using Unit Circle Trigonometry, we know that cos    1 .
Negative x-axis:     2 n   ( 2 n  1 )  , where n is an integer
Answer:
1f.
 

 2 n  ,   ( 2 n  1 )  , where n is an integer
2
2 sin 2   3 sin   1  0
Back to Problem 1.
NOTE: This equation is quadratic in sin  . Factoring this equation, we
obtain the following.
2 sin 2   3 sin   1  0  ( sin   1) ( 2 sin   1)  0 
sin   1  0 , 2 sin   1  0
NOTE: There are two equations to solve:
1.
sin   1  0
and 2.
2 sin   1  0
The solutions of the equation sin   1  0 :
sin   1  0  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.
 
sin
    1.
NOTE: Using Unit Circle Trigonometry, we know that
 2
Negative y-axis:   

 2 n  , where n is an integer
2
The solutions of the equation 2 sin   1  0 :
2 sin   1  0  2 sin    1  sin   
1
2
NOTE: Sine is negative in the third and fourth quadrants.
To find the reference angle  ' :
sin   
7
 2 n  , where n is an integer
6
III:
 
IV:
  
Answer:
1
1
1

 sin  ' 
  '  sin  1 
2
2
2
6

 2 n
6
  


7
 2 n ,   
 2 n ,  
 2 n  , where
2
6
6
n is an integer
1g.
6 cos 2   7 cos   5  0
Back to Problem 1.
NOTE: This equation is quadratic in cos  . Factoring this equation, we
obtain the following.
6 cos 2   7 cos   5  0  ( 2 cos   1) ( 3 cos   5 )  0 
2 cos   1  0 , 3 cos   5  0
NOTE: There are two equations to solve:
1.
2 cos   1  0
and 2.
3 cos   5  0
The solutions of the equation 2 cos   1  0 :
2 cos   1  0  2 cos    1  cos   
1
2
NOTE: Cosine is negative in the second and third quadrants.
To find the reference angle  ' :
cos   
1
1
1

 cos  ' 
  '  cos  1 
2
2
2
3
II:
 
2
 2 n  , where n is an integer
3
III:
 
4
 2 n
3
The solutions of the equation 3 cos   5  0 :
3 cos   5  0  3 cos   5  cos  
5
3
5
does not have any solutions. We know that
3
the cosine of any angle must be a number in the closed interval [  1, 1 ] .
NOTE: The equation cos  
Thus,  1  cos   1 for all angles  . Since
exist an angle  such that cos  
Answer:
1h.
 
5
 1 , then there does not
3
5
.
3
2
4
 2 n ,  
 2 n  , where n is an integer
3
3
sin 2   24  11sin 
Back to Problem 1.
NOTE: This equation is quadratic in sin  .
sin 2   24  11sin   sin 2   11sin   24  0 
( sin   3 ) ( sin   8 )  0  sin   3  0 , sin   8  0
NOTE: There are two equations to solve:
1.
sin   3  0
and 2.
sin   8  0
The solutions of the equation sin   3  0 :
sin   3  0  sin   3
NOTE: The equation sin   3 does not have any solutions. We know that
the sine of any angle must be a number in the closed interval [  1, 1 ] . Thus,
 1  sin   1 for all angles  . Since 3  1 , then there does not exist an
angle  such that sin   3 .
The solutions of the equation sin   8  0 :
sin   8  0  sin   8
NOTE: The equation sin   8 does not have any solutions. We know that
the sine of any angle must be a number in the closed interval [  1, 1 ] . Thus,
 1  sin   1 for all angles  . Since 8  1 , then there does not exist an
angle  such that sin   8 .
Answer:
1i.
No solution
tan 2 x  5 tan x  6  0
Back to Problem 1.
NOTE: This equation is quadratic in tan x .
tan 2 x  5 tan x  6  0  ( tan x  1) ( tan x  6 )  0 
tan x  1  0 , tan x  6  0
NOTE: There are two equations to solve:
1.
tan x  1  0
and 2.
tan x  6  0
The solutions of the equation tan x  1  0 :
tan x  1  0  tan x   1
NOTE: Tangent is negative in the second and fourth quadrants.
To find the reference angle x ' :
tan x   1  tan x '  1  x '  tan  1 1 

4
3
 2 n  , where n is an integer
4
II:
x 
IV:
x  

 2 n
4
NOTE: The two solutions of x  
written as the one solution x  

3
 2 n  and x 
 2 n  may be
4
4

 n  , where n is an integer.
4
The solutions of the equation tan x  6  0 :
tan x  6  0  tan x  6
NOTE: Tangent is positive in the first and third quadrants.
To find the reference angle x ' :
tan x  6  tan x '  6  x '  tan  1 6
I:
x  tan  1 6  2 n  , where n is an integer
III:
x    tan  1 6  2 n   tan  1 6  ( 2 n  1) 
1
NOTE: The two solutions of x  tan 6  2 n  and
x  tan  1 6  ( 2 n  1)  may be written as the one solution
x  tan  1 6  n  , where n is an integer.
Answer:

3
 2 n , x 
 2 n  , x  tan  1 6  2 n  ,
4
4
x  tan  1 6  ( 2 n  1)  , where n is an integer
x  
OR x  

 n  , x  tan  1 6  n  , where n is an
4
integer
1j.
3 sec 2   sec   14  0
Back to Problem 1.
NOTE: This equation is quadratic in sec  .
3 sec 2   sec   14  0  ( sec   2 ) ( 3 sec   7 )  0 
sec   2  0 , 3 sec   7  0
NOTE: There are two equations to solve:
1.
sec   2  0
and 2.
3 sec   7  0
The solutions of the equation sec   2  0 :
sec   2  0  sec   2  cos  
1
2
NOTE: Cosine is positive in the first and fourth quadrants.
To find the reference angle  ' :
cos  
1
1
1

 cos  ' 
  '  cos  1 
2
2
2
3

 2 n  , where n is an integer
3
I:
 
IV:
  

 2 n
3
The solutions of the equation 3 sec   7  0 :
3 sec   7  0  3 sec    7  sec   
7
3
 cos   
3
7
NOTE: Cosine is negative in the second and third quadrants.
To find the reference angle  ' :
cos   
II:
    cos  1
3
3
3
 cos  ' 
  '  cos  1
7
7
7
3
3
 2 n    cos  1  ( 2 n  1 )  , where n is an
7
7
integer
III:
    cos  1
Answer:
3
3
 2 n   cos  1  ( 2 n  1 ) 
7
7


 2 n ,  
 2 n ,
3
3
3
3
   cos  1  ( 2 n  1 )  ,   cos  1  ( 2 n  1 )  ,
7
7
  
where n is an integer
Solution to Problems on the Pre-Exam:
21.
Back to Page 1.
Find all the exact solutions (in degrees) of the equation
( cos   1 ) ( csc   2 )  0 . (8 pts.) Put a box around your answer.
NOTE: There are two equations to solve:
1.
cos   1  0
and 2.
csc   2  0
The solutions of the equation cos   1  0 :
cos   1  0  cos    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
cosine function.
NOTE: Using Unit Circle Trigonometry, we know that cos 180    1 .
Negative x-axis:   180   n  360  , where n is an integer
The solutions of the equation csc   2  0 :
csc   2  0  csc    2  sin   
1
2
NOTE: Sine is negative in the third and fourth quadrants.
To find the reference angle  ' :
sin   
1
1
1
 sin  ' 
  '  sin  1  30 
2
2
2
III:
  210   n  360  , where n is an integer
IV:
   30   n  360 