Download Pre-Class Problems 16 for Tuesday, November 18 These are the

Pre-Class Problems 16 for Tuesday, November 18
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.
1.
Find all the exact solutions for the following equations.
a. cos   0
b.
5 sin   8  3
c.
6 sec   5  1
d.
sin x  0
e.
tan    4
f.
8 sec   15   3
Objective of the following problems: To find all the solutions to a trigonometric
equation in a specified interval.
3
2a. Find all the exact solutions (in radians) of the equation cos 3 x  
in
2
the interval [   ,  ] . Put a box around your answer(s).
2b.
Find all the exact solutions (in degrees) of the equation 2 sin
in the interval [  900  , 1620  ] .
2c.
Find all the exact solutions (in radians)
3 cot ( 2    )  3  0 in the interval [ 0 , 2  ] .
of
2d.
Find all the exact solutions (in degrees) of the equation
5 sec ( 4   80  )  6  4 in the interval [  540  , 270  ] .


2
2  0
the
equation
2e.
Find
all
the exact solutions (in radians) of
3 

2 sin  x 
  3  0 in the interval [  3 , 5  ] .
4 

2f.
Find all the exact solutions (in degrees) of the equation sin 6   0 in the
interval [  90  , 180  ] .
2g.
Find
all
the exact solutions (in radians) of
2 

7 sec  5 
  11  4 in the interval [   , 2  ] .
3 

the
the
equation
equation
Additional problems available in the textbook: Page 231 … 39 - 44, 45. Examples
7 and 8 on page 229.
Solutions:
1a.
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:
1b.
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:  
1c.
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
1d.
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.
1e.
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
1f.
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
2a.
cos 3 x  
2
 2 n  , where n is an integer
3
2
2
 2 n  ,    cos  1  2 n  , where n is an integer
3
3
3
, [  ,  ]
2
Back to Problem 2.
Let u  3 x . Then cos 3 x  
3
3
 cos u  
2
2
NOTE: Cosine is negative in the second and third quadrants.
To find the reference angle u ' :
cos u  
3
 cos u ' 
2
3
3

 u '  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 u    u '   
.
6
6
6
6
All solutions for u in II: u 
5
 2 n  , where n is an integer
6
5
 2 n  and u  3 x 
6
5
5
2 n
3x 
 2 n  x 

, where n is an integer.
6
18
3
All solutions for x in II: u 
Simplifying, we have that x 
=
5
2 n
5   12 n 
5
12 n 


=
=
18
3
18
18
18
(12 n  5 ) 
, where n is an integer
18
The solutions for x 
n  0: x 
( 12 n  5 ) 
which are in the interval [   ,  ] :
18
( 0  5)

18
5
18
n  1: x 
( 12  5 ) 

18
n  2: x 
( 24  5 ) 
29 

18
18
NOTE:
17 
18
29 
18 

  . So, this solution is not in the interval [   ,  ] .
18
18
n   1: x 
(  12  5 ) 
 7

18
18
n   2: x 
(  24  5 ) 
 19 

18
18
NOTE:
[  ,  ].

19 
18 
 
   . So, this solution is not in the interval
18
18
NOTE: The one solution in the third quadrant that is in the interval [ 0 , 2  )

6

7



is given by u    u '   
.
6
6
6
6
All solutions for u in III: u 
7
 2 n  , where n is an integer
6
7
 2 n  and u  3 x 
6
7
7
2 n
3x 
 2 n  x 

, where n is an integer.
6
18
3
All solutions for x in III: u 
Simplifying, we have that x 
=
7
2 n
7
12 n 
7   12 n 


=
=
18
3
18
18
18
( 12 n  7 ) 
, where n is an integer
18
The solutions for x 
( 12 n  7 ) 
which are in the interval [   ,  ] :
18
n  0: x 
( 0  7 )

18
7
18
n  1: x 
( 12  7 ) 

18
19 
18
NOTE:
19 
18 

  . So, this solution is not in the interval [   ,  ] .
18
18
n   1: x 
(  12  7 ) 
 5


18
18
n   2: x 
(  24  7 ) 
 17 


18
18
n   3: x 
(  36  7 ) 
 29 
29 

 
18
18
18
NOTE:

[  ,  ].
Answer: 

5
18

17 
18
29 
18 
 
   . So, this solution is not in the interval
18
18
5
7
17 
,
,
18
18
18
2b.
2 sin


2
2  0 , [  900  , 1620  ]
Let u 



. Then 2 sin
2
2
2 sin u 
2  sin u 
2  0  2 sin u 
Back to Problem 2.
2  0 
2
2
NOTE: Sine is positive in the first and second quadrants.
To find the reference angle u ' :
sin u 
2
 sin u ' 
2
2
2
 u '  sin  1
 45 
2
2
NOTE: The one solution in the first quadrant that is in the interval
[ 0  , 360  ) is given by u  u '  45  .
All solutions for u in I: u  45   n  360  , where n is an integer
All solutions for  in I: u  45   n  360  and u 


2

 45   n  360     90   n  720  , where n is an integer.
2
The solutions for   90   n  720  which are in the interval
[  900  , 1620  ] :
n  0 :   90   0  
n  1 :   90   720  
n  2 :   90   1440  
90 
810 
1530 
n  3 :   90   2160   2250 
NOTE: 2250   1620  . So, this solution is not in the interval
[  900  , 1620  ] .
n   1 :   90   720    630 
n   2 :   90   1440    1350 
 1350    900  .
NOTE:
[  900  , 1620  ] .
So, this solution is not in the interval
NOTE: The one solution in the second quadrant that is in the interval
[ 0  , 360  ) is given by u  180   u '  180   45   135  .
All solutions for u in II: u  135   n  360  , where n is an integer
All solutions for  in II: u  135   n  360  and u 


2

 135   n  360     270   n  720  , where n is an integer.
2
The solutions for   270   n  720  which are in the interval
[  900  , 1620  ] :
n  0 :   270   0  
270 
n  1 :   270   720  
990 
n  2 :   270   1440   1710 
NOTE: 1710   1620  . So, this solution is not in the interval
[  900  , 1620  ] .
n   1 :   270   720  
 450 
n   2 :   270   1440    1170 
 1170    900  .
NOTE:
[  900  , 1620  ] .
So, this solution is not in the interval
Answer:  630  ,  450  , 90  , 270  , 810  , 990  , 1530 
2c.
3 cot ( 2    ) 
3  0 , [ 0 , 2 ]
Let u  2    . Then 3 cot ( 2    ) 
3 cot u 
3  0  3 cot u  
tan u  
3
 tan u  
3
Back to Problem 2.
3  0 
3  cot u  
3

3
3
NOTE: Tangent is negative in the second and fourth quadrants.
To find the reference angle u ' :
tan u  
3  tan u ' 
3  u '  tan  1 3 

3
NOTE:
The one solution in the second quadrant that is in the interval

3

2



[ 0 , 2  ) is given by u    u '   
.
3
3
3
3
All solutions for u in II: u 
2
 2 n  , where n is an integer
3
2
 2 n  and u  2    
3
2


2   
 2 n  2   
 2 n    
 n ,
3
3
6
where n is an integer.
All solutions for  in II: u 
Simplifying, we have that   
=


6 n
   6 n
 n = 

=
6
6
6
6
( 6 n  1) 
, where n is an integer
6
The solutions for  
n  0:  
NOTE: 
( 6 n  1) 
which are in the interval [ 0 , 2  ] :
6
( 0  1) 



 
6
6
6

 0 . So, this solution is not in the interval [ 0 , 2  ] .
6
n  1:  
( 6  1) 

6
n  2:  
( 12  1 ) 

6
n  3:  
(18  1) 
17 

6
6
NOTE:
5
6
11
6
17 
12 

 2  . So, this solution is not in the interval [ 0 , 2  ] .
6
6
n   1:  
(  6  1) 
 7


6
6

7
6
NOTE: 
NOTE:
7
 0 . So, this solution is not in the interval [ 0 , 2  ] .
6
The one solution in the fourth quadrant that is in the interval
(  2  , 0 ] is given by u   u '  
All solutions for u in IV: u  

.
3

 2 n  , where n is an integer
3

 2 n  and u  2    
3

4
2
2    
 2 n  2   
 2 n    
 n ,
3
3
3
All solutions for  in IV: u  
where n is an integer.
Simplifying, we have that   
2
2
3n
 n = 

=
3
3
3
 2  3 n
( 3n  2 )
=
, where n is an integer
3
3
The solutions for  
n  0:  
NOTE: 
( 3n  2 )
which are in the interval [ 0 , 2  ] :
3
( 0  2 )
 2
2

 
3
3
3
2
 0 . So, this solution is not in the interval [ 0 , 2  ] .
3
n  1:  
( 3  2 )

3

3
n  2:  
( 6  2 )

3
4
3
n  3:  
NOTE:
( 9  2 )
7

3
3
7
6

 2  . So, this solution is not in the interval [ 0 , 2  ] .
3
3
n   1:  
2d.
(  3  2 )
 5


3
3

5
3
NOTE: 
5
 0 . So, this solution is not in the interval [ 0 , 2  ] .
3
Answer:
 5  4  11
,
,
,
3 6 3
6
5 sec ( 4   80  )  6  4 , [  540  , 270  ]
Back to Problem 2.
Let u  4   80  . Then 5 sec ( 4   80  )  6  4 
5 sec u  6  4  5 sec u  10  sec u  2  cos u 
1
2
NOTE: Cosine is positive in the first and fourth quadrants.
To find the reference angle u ' :
cos u 
1
1
1
 cos u ' 
 u '  cos  1  60 
2
2
2
NOTE: The one solution in the first quadrant that is in the interval
[ 0  , 360  ) is given by u  u '  60  .
All solutions for u in I: u  60   n  360  , where n is an integer
All solutions for  in I: u  60   n  360  and u  4   80  
4  80   60   n  360   4   140   n  360     35   n  90  ,
where n is an integer.
The solutions for   35   n  90  which are in the interval
[  540  , 270  ] :
n  0 :   35   0  
35 
n  1 :   35   90  
125 
n  2 :   35   180  
215 
n  3 :   35   270   305 
NOTE: 305   270  . So, this solution is not in the interval
[  540  , 270  ] .
n   1 :   35   90  
 55 
n   2 :   35   180  
 145 
n   3 :   35   270  
 235 
n   4 :   35   360  
 325 
n   5 :   35   450  
 415 
n   6 :   35   540  
 505 
n   7 :   35   630    595 
 595    540  .
NOTE:
[  540  , 270  ] .
So, this solution is not in the interval
NOTE: The one solution in the fourth quadrant that is in the interval
(  360  , 0  ] is given by u   u '   60  .
All solutions for u in IV: u   60   n  360  , where n is an integer
All solutions for  in IV: u   60   n  360  and u  4   80  
4   80    60   n  360   4   20   n  360     5   n  90  ,
where n is an integer.
The solutions for   5   n  90  which are in the interval
[  540  , 270  ] :
n  0 :   5  0 
5
n  1 :   5   90  
95 
n  2 :   5   180  
185 
n  3 :   5   270   275 
NOTE: 275   270  . So, this solution is not in the interval
[  540  , 270  ] .
n   1 :   5   90  
 85 
n   2 :   5   180  
 175 
n   3 :   5   270  
 265 
n   4 :   5   360  
 355 
n   5 :   5   450  
 445 
n   6 :   5   540  
 535 
n   7 :   5   630    625 
 625    540  .
NOTE:
[  540  , 270  ] .
So, this solution is not in the interval
Answer:  535  ,  505  ,  445  ,  415  ,  355  ,  325  ,  265  ,  235  ,
 175  ,  145  ,  85  ,  55  , 5  , 35  , 95  , 125  , 185  , 215 
2e.
3 

2 sin  x 
 
4 

Let u  x 
2 sin u 
3  0 , [  3 , 5  ]
3 
3

2
sin
x


 
. Then
4 
4

3  0  2 sin u  
Back to Problem 2.
3  0 
3  sin u  
3
2
NOTE: Sine is negative in the third and fourth quadrants.
To find the reference angle u ' :
sin u  
3
 sin u ' 
2
3
3

 u '  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 u    u '   
.
3
3
3
3
4
 2 n  , where n is an integer
3
All solutions for u in III: u 
3
4

 2 n  and u  x 
4
3
3
4
7
x 

 2 n  x 
 2 n  , where n is an integer.
4
3
12
All solutions for x in III: u 
NOTE:
4
3
16 
9
7




3
4
12
12
12
7
7
24 n 
 2 n =

=
12
12
12
Simplifying, we have that x 
7   24 n 
( 24 n  7 ) 
=
, where n is an integer
12
12
The solutions for x 
( 24 n  7 ) 
which are in the interval [  3 , 5  ] :
12
n  0: x 
( 0  7 )

12
7
12
n  1: x 
( 24  7 ) 

12
31
12
n  2: x 
( 48  7 ) 

12
55 
12
n  3: x 
( 72  7 ) 
79 

12
12
79 
60 

 5 .
12
12
[  3 , 5  ] .
NOTE:
So, this solution is not in the interval
n   1: x 
(  24  7 ) 
 17 


12
12

17 
12
n   2: x 
(  48  7 ) 
 41


12
12

41
12
41
36 
 
  3  . So, this solution is not in the interval
12
12
[  3 , 5  ] .
NOTE:

NOTE:
The one solution in the fourth quadrant that is in the interval
(  2  , 0 ] is given by u   u '  
All solutions for u in IV: u  

.
3

 2 n  , where n is an integer
3
3


 2 n  and u  x 
4
3
3

13 
x 
 
 2 n  x  
 2 n  , where n is an integer.
4
3
12
All solutions for x in IV: u  
NOTE: 

3
4
9
13 

 

 
3
4
12
12
12
13 
13 
24 n 
 2 n = 

=
12
12
12
 13   24 n 
( 24 n  13 ) 
=
, where n is an integer
12
12
Simplifying, we have that x  
The solutions for x 
n  0: x 
( 24 n  13 ) 
which are in the interval [  3 , 5  ] :
12
( 0  13 ) 
 13 


12
12

13 
12
n  1: x 
( 24  13 ) 

12
11
12
n  2: x 
( 48  13 ) 

12
35 
12
n  3: x 
( 72  13 ) 

12
59 
12
n  4: x 
( 96  13 ) 
83 

12
12
83 
60 

 5 .
12
12
[  3 , 5  ] .
NOTE:
So, this solution is not in the interval
n   1: x 
(  24  13 ) 
 37 


12
12

37 
12
n   2: x 
(  48  13 ) 
 61


12
12

61
12
61
36 
 
  3 . So, this solution is not in the interval
12
12
[  3 , 5  ] .
NOTE:

Answer: 
2f.
37 
17 
13  7  11 31 35  55  59 
,
,
,
,
,
,
,
,
12
12
12 12 12 12 12 12 12
sin 6   0 , [  90  , 180  ]
Let u  6  . Then sin 6   0  sin u  0
Back to Problem 2.
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 u  0  .
All solutions for u on the positive x-axis:
where n is an integer
u  0   n  360   n  360  ,
NOTE: Using Unit Circle Trigonometry, we know that sin 180   0 . Thus,
the one solution, which lies on the negative x-axis, is u  180  .
All solutions for u on the negative x-axis: u  180   n  360  , where n is an
integer
NOTE: These two solutions of u  0   n  360  and u  180   n  360  ,
where n is an integer, may be written as u  0   n 180   n 180  , where
n is an integer.
All solutions for  : u  n 180  and u  6   6   n 180  
  n  30  , where n is an integer.
The solutions for   n  30  which are in the interval [  90  , 180  ] :
n  0:  
0
n  1:  
30 
n  2:  
60 
n  3:  
90 
n  4:  
120 
n  5:  
150 
n  6:  
180 
n   1:  
 30 
n   2:  
 60 
n   3:  
 90 
Answer:  90  ,  60  ,  30  , 0  , 30  , 60  , 90  , 120  , 150  , 180 
2g.
2 

7 sec  5 
  11  4 , [   , 2  ]
3 

Let u  5 
Back to Problem 2.
2 
2

  11  4 
. Then 7 sec  5 
3 
3

7 sec u  11  4  7 sec u   7  sec u   1  cos u   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 . Thus,
the one solution, which lies on the negative x-axis, is u   .
All solutions for u on the negative x-axis: u    2 n  , where n is an
integer
2

3
2
5

2 n
5 
   2 n   5 
 2 n   

,
3
3
3
5
where n is an integer.
All solutions for  : u    2 n  and u  5 
Simplifying, we have that  

2 n
5
6 n


=
=
15
15
3
5
5  6 n 
( 6 n  5)
=
, where n is an integer
15
15
The solutions for  
( 6 n  5)
which are in the interval [   , 2  ] :
15
n  0:  
( 0  5)
5


15
15
n  1:  
( 6  5)

15
n  2:  
( 12  5 ) 

15
17 
15
n  3:  
( 18  5 ) 

15
23 
15
n  4:  
( 24  5 ) 

15
29 
15
n  5:  
( 30  5 ) 
35 
7


15
15
3
NOTE:

3
11
15
7
6

 2  . So, this solution is not in the interval [   , 2  ] .
3
3
n   1:  
(  6  5)

15
n   2:  
(  12  5 ) 

15


15

7
15
n   3:  
(  18  5 ) 

15

13 
15
n   4:  
(  24  5 ) 

15

19 
15
NOTE:

[   , 2 ] .
Answer: 
19 
15 
 
   . So, this solution is not in the interval
15
15
13
7
  11 17  23 29 
,
, , ,
,
,
,
15
15
15 3 15 15 15 15
Solution to Problems on the Pre-Exam:
25.
Back to Page 1.
3
in the
2
interval [  180  , 180  ] . (9 pts.) Put a box around your answer(s).
Find all the solutions (in degrees) to the equation cos 3 x  
Let u  3 x . Then cos 3 x  
3
3
 cos u  
2
2
NOTE: Cosine is negative in the second and third quadrants.
To find the reference angle u ' :
cos u  
3
 cos u ' 
2
3
3
 u '  cos  1
 30 
2
2
NOTE: The one solution in the second quadrant that is in the interval
[ 0 , 360  ) is given by u  180   u '  180   30   150  .
All solutions for u in II: u  150   n  360  , where n is an integer
All solutions for x in II: u  150   n  360  and u  3 x 
3 x  150   n  360   x  50   n 120   , where n is an integer.
The solutions for x  50   n 120  which are in the interval [  180  , 180  ] :
n  0 : x  50 
n  1 : x  50   120   170 
n  2 : x  50   240   290   180 
n   1 : x  50   120    70 
n   2 : x  50   240    190    180 
NOTE: The one solution in the third quadrant that is in the interval
[ 0 , 360  ) is given by u  180   u '  180   30   210  .
All solutions for u in III: u  210   n  360  , where n is an integer
All solutions for x in III: u  210   n  360  and u  3 x 
3 x  210   n  360   x  70   n 120   , where n is an integer.
n  0 : x  70 
n  1 : x  70   120   190   180 
n   1 : x  70   120    50 
n   2 : x  70   240    170 
n   3 : x  70   360    290    180 