Lesson 10 - Solving trigonometric equations PDF

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LESSON 10 SOLVING TRIGONOMETRIC EQUATIONS

Equations which are solved in this lesson (Click on the number or equation): 3 cos    2sin   2  0 2. 1. 2 3sec x  4  2 3 cot   1  0 4. 3. 5.

  14 sin    12  5 3

6.

  tan  5    1 3 

7.

csc ( 4   280  )  

8.

cos 6  0

9.

sec 2   2  0

10.

4sin 2   9  12

11.

cos  ( csc   2 )  0

12.

3 sin x tan x  sin x

13.

( cos   1) ( sin   1)  0 6 sin 2   7 sin   5  0

14.

2 cos 2   cos   1  0

16.

cos 2   24  11cos 

15.

2 3

Additional examples worked in this lesson: Example Find all the approximate solutions (in degrees) between  360  and 5 540  for the equation tan    . 4 Example Find all the approximate solutions (in degrees) between  270  and 630  for the equation 3sec 2  7  0. Example Find all the approximate solutions (in degrees) between  360  and 360  for the equation 9sin ( 3  150 )  4  0. Example Find all the exact and approximate solutions to the equation 2 tan   6 tan   11  0 .

Examples Find all the exact solutions for the following equations.

1.

cos   

3 2

3

First, determine where the solutions  will occur. Since 

is not the 2 minimum negative number for the cosine function, then the solutions do not occur at the coordinate axes. Thus, the solutions occur in two of the four quadrants. Since cosine is negative in the II and III quadrants, the solutions for this equation occur in those quadrants. Find the reference angle  ' for the solutions  :

cos   

3 2



cos  ' 

3 2



 '  cos  1

3 2



 6

The solutions in the II quadrant: The one solution in the II quadrant, that is 5   between 0 and 2 , is     . Now, all the other solutions in 6 6 the II quadrant are coterminal to this one solution. Thus, all the solutions in 5  2 n , where n is an integer. the II quadrant are given by   6 Thus, some of the solutions in the II quadrant that are obtained by using 5    2 n , where n is an integer, are: 6

n  0:

 

5 5  0  6 6

n  1:

 

5 5 12 17   2    6 6 6 6

n  2:

 

5 5 24 29  4    6 6 6 6

n  3:

 

5 5 36 41  6    6 6 6 6

n  4:

 

5 5 48 53  8    6 6 6 6

: : : n   1:

 

n   2:  

5 5 12 7  2     6 6 6 6 5 5 24 19  4     6 6 6 6

 

5 5 36 31  6     6 6 6 6

n   4:  

5 5 48 43  8     6 6 6 6

n   3:

: : : The solutions in the III quadrant: The one solution in the III quadrant, that is 7   between 0 and 2 , is     . Now, all the other solutions in the 6 6 III quadrant are coterminal to this one solution. Thus, all the solutions in the 7  2 n  , where n is an integer. III quadrant are given by   6 Thus, some of the solutions in the III quadrant that are obtained by using 7    2 n , where n is an integer, are: 6

n  0:

 

7 7 0  6 6

n  1:

 

7 7 12 19  2    6 6 6 6

n  2:

 

7 7 24 31  4    6 6 6 6

n  3:

 

7 7 36 43  6    6 6 6 6

n  4:

 

7 7 48 55  8    6 6 6 6

: : :  

7 7 12 5  2     6 6 6 6

n   2:  

7 7 24 17  4     6 6 6 6

 

7 7 36 29  6     6 6 6 6

n   4:  

7 7 48 41  8     6 6 6 6

n   1:

n   3:

: : :

Answers:  

2.

2sin  

5 7  2n  ;    2 n , where n is an integer 6 6

2 0

First, do the necessary algebra to isolate sin  on one side of the equation:

2sin  

2 0 

2sin  

2 

2 2

sin   2

Now, determine where the solutions  will occur. Since

is not the 2 maximum positive number for the sine function, then the solutions do not occur at the coordinate axes. Thus, the solutions occur in two of the four quadrants. Since sine is positive in the I and II quadrants, the solutions for this equation occur in those quadrants. Find the reference angle  ' for the solutions  :

sin  

2



2

sin  ' 

2 2



 '  sin  1

2 2



 4

The solutions in the I quadrant: The one solution in the I quadrant, that is between 0 and 2 , is  



. Now, all the other solutions in the I quadrant 4 are coterminal to this one solution. Thus, all the solutions in the I quadrant are given by  



4

 2 n  , where n is an integer.

Thus, some of the solutions in the I quadrant that are obtained by using

 



4

 2 n , where n is an integer, are:

n  0:

 

n  1:

 

n  2:

 

n  3:

 

n  4:

 

 4

 4

 4

 4

 4

 0

 4

 2 

 4   6   8 

 4

 4

 4

 4



8 9  4 4



16 17  4 4



24 25   4 4



32 33  4 4



8 7   4 4



16 15   4 4



24 23   4 4



32 31   4 4

: : : n   1:

 

n   2:  

n   3:

 

n   4:  

 4

 4

 4

 4

 2 

 4   6   8 

 4

 4

 4

 4

: : :

The solutions in the II quadrant: The one solution in the II quadrant, that is 3   between 0 and 2 , is     . Now, all the other solutions in 4 4 the II quadrant are coterminal to this one solution. Thus, all the solutions in 3  2 n , where n is an integer. the II quadrant are given by   4 Thus, some of the solutions in the II quadrant that are obtained by using 3    2 n , where n is an integer, are: 4

n  0:

 

3 3 0  4 4

n  1:

 

3 3 8 11  2    4 4 4 4

n  2:

 

3 3 16 19  4    4 4 4 4

n  3:

 

3 3 24 27  6    4 4 4 4

n  4:

 

3 3 32 35  8    4 4 4 4

: : : n   1:

 

n   2:  

3 3 8 5  2     4 4 4 4 3 3 16 13  4     4 4 4 4

 

3 3 24 21  6     4 4 4 4

n   4:  

3 3 32 29  8     4 4 4 4

n   3:

: : : Answers:  

3.

 4

 2 n ;  

3  2 n , where n is an integer 4

3 cot   1  0 First, do the necessary algebra to isolate cot  on one side of the equation:

3 cot   1  0 

3 cot    1 

cot   

1 3

Now, take the reciprocal of both sides of this equation to obtain the equation tan    3 . Now, determine where the solutions  will occur. Since tangent is negative in the II and IV quadrants, the solutions for this equation occur in those quadrants. Find the reference angle  ' for the solutions  :

tan    3 

tan  ' 

3 

 '  tan  1

3 

 3

The solutions in the II quadrant: The one solution in the II quadrant, that is 2   between 0 and 2 , is     . Now, all the other solutions in 3 3 the II quadrant are coterminal to this one solution. Thus, all the solutions in 2  2 n  , where n is an integer. the II quadrant are given by   3 Thus, some of the solutions in the II quadrant that are obtained by using 2    2 n , where n is an integer, are: 3

n  0:

 

2 2 0  3 3

n  1:

 

2 2 6 8  2    3 3 3 3

n  2:

 

2 2 12 14  4    3 3 3 3

n  3:

 

2 2 18 20  6    3 3 3 3

n  4:

 

2 2 24 26  8    3 3 3 3

: : : n   1:

 

2 2 6 4  2     3 3 3 3

n   2:

 

2 2 12 10  4     3 3 3 3

n   3:

 

2 2 18 16  6     3 3 3 3

n   4:

 

2 2 24 22  8     3 3 3 3

: : : The solutions in the IV quadrant: The one solution in the IV quadrant, that is 5   between 0 and 2 , is   2   . Now, all the other solutions in 3 3 the IV quadrant are coterminal to this one solution. Thus, all the solutions in 5  2 n  , where n is an integer. the IV quadrant are given by   3 Thus, some of the solutions in the IV quadrant that are obtained by using 5  2 n , where n is an integer, are:   3

n  0:

 

5 5  0 3 3

n  1:

 

5 5 6 11  2    3 3 3 3

n  2:

 

5 5 12 17  4    3 3 3 3

n  3:

 

5 5 18 23  6    3 3 3 3

n  4:

 

5 5 24 29  8    3 3 3 3

: : : n   1:

 

5 5 6   2     3 3 3 3

n   2:

 

5 5 12 7  4     3 3 3 3

n   3:

 

5 5 18 13  6     3 3 3 3

n   4:

 

5 5 24 19  8     3 3 3 3

: : : Answers:  

5 2  2 n , where n is an integer  2n  ;   3 3

NOTE: Since the period of the tangent function is  , then these two answers 2  n , where n is an integer. may also be written as the one answer   3

4.

3sec x  4  2

First, do the necessary algebra to isolate sec x on one side of the equation:

3sec x  4  2 

3sec x  6 

sec x  2

Now, take the reciprocal of both sides of this equation to obtain the equation 1 cos x  . 2

1 is not the 2 maximum positive number for the cosine function, then the solutions do not occur at the coordinate axes. Thus, the solutions occur in two of the four quadrants. Since cosine is positive in the I and IV quadrants, the solutions for this equation occur in those quadrants. Now, determine where the solutions x will occur.

Since

Find the reference angle x ' for the solutions x :

cos x 

1  2

cos x ' 

1  2

x'  cos  1

 1  2 3

The solutions in the I quadrant: The one solution in the I quadrant, that is between 0 and 2 , is x 



. Now, all the other solutions in the I quadrant 3 are coterminal to this one solution. Thus, all the solutions in the I quadrant are given by x 



3

 2 n  , where n is an integer.

The solutions in the IV quadrant: The one solution in the IV quadrant, that is 5   between 0 and 2 , is x  2  . Now, all the other solutions in 3 3 the IV quadrant are coterminal to this one solution. Thus, all the solutions in 5  2 n  , where n is an integer. the IV quadrant are given by x  3

Answers: x 

 3

 2 n ; x 

5  2 n , where n is an integer 3

5.

  14 sin    12  5 3 First, we must get the angle argument for the trigonometric function to be a single variable.

So, let u 



. Thus, we obtain the equation 3 14 sin u  12  5 . Now, do the necessary algebra to isolate sin u on one side of the equation:

14 sin u  12  5 

14 sin u   7 

sin u  

1 2

1 is not the 2 minimum negative number for the sine function, then the solutions do not occur at the coordinate axes. Thus, the solutions occur in two of the four quadrants. Since sine is negative in the III and IV quadrants, the solutions for this equation occur in those quadrants. Now, determine where the solutions u will occur. Since 

Find the reference angle u ' for the solutions u :

sin u  

1  2

sin u ' 

1  2

u '  sin  1

1   2 6

The solutions in the III quadrant: The one solution u in the III quadrant, that 7   is between 0 and 2 , is u    . Now, all the other solutions u 6 6 in the III quadrant are coterminal to this one solution. Thus, all the solutions 7  2 n  , where n is an integer. u in the III quadrant are given by u  6 These are the solutions for u. Now, find the solutions for  . Since u 



3

7 7   2 n , then   2 n . Now, multiply both sides of 6 3 6 7  6 n  , where n is this equation by 3 in order to solve for  . Thus,   2 an integer. and u 

Thus, some of the solutions in the III quadrant that are obtained by using 7  6 n , where n is an integer, are:   2

n  0:

n  1:

n  2:

n  3:

n  4:

7 7  0  2 2 7   NOTE: is in the III quadrant. 3 6

 

7 7 12 19  6    2 2 2 2 7  19    2  NOTE: is in the III quadrant. 3 6 6

 

7 7 24 31  12    2 2 2 2 7  31     4 is in the III quadrant. NOTE: 3 6 6

 

7 7 36  43  18    2 2 2 2 43 7    6  NOTE: is in the III quadrant. 3 6 6

 

7 7 48 55  24    2 2 2 2 7  55   8  NOTE: is in the III quadrant. 3 6 6

 

: : : n   1:



7 7 12 5  6     2 2 2 2

 NOTE:

n   2:

n   3:

n   4:

3

 

5 is in the III quadrant. 6

7 7 24 17  12     2 2 2 2 17 5      2  NOTE: is in the III quadrant. 3 6 6

 

7 7 36 29  18     2 2 2 2 29  5      4  NOTE: is in the III quadrant. 3 6 6

 

7 7 48 41  24     2 2 2 2 41 5  6       NOTE: is in the III quadrant. 3 6 6

 

: : : The solutions in the IV quadrant: The one solution u in the IV quadrant, that  11  is between 0 and 2 , is u  2  . Now, all the other solutions 6 6 u in the IV quadrant are coterminal to this one solution. Thus, all the 11  2 n , where n is an solutions u in the IV quadrant are given by u  6 integer. These are the solutions for u. Now, find the solutions for  . Since 11   11  2 n , then u    2 n  . Now, multiply both and u  3 6 3 6 11  6 n , sides of this equation by 3 in order to solve for  . Thus,   2 where n is an integer.

Thus, some of the solutions in the IV quadrant that are obtained by using 11    6 n , where n is an integer, are: 2

n  0:

n  1:

n  2:

n  3:

n  4:

11 11  0  2 2  11  NOTE: is in the IV quadrant. 3 6

 

11 11 12 23  6    2 2 2 2 11  23   2  NOTE: is in the IV quadrant. 3 6 6

 

11 11 24 35  12    2 2 2 2 11  35 4     NOTE: is in the IV quadrant. 3 6 6

 

11 11 36 47  18    2 2 2 2  47 11   6  is in the IV quadrant. NOTE: 3 6 6

 

11 11 48 59  24    2 2 2 2 11  59   8  NOTE: is in the IV quadrant. 3 6 6

 

: : : n   1:

 

11 11 12   6     2 2 2 2

 NOTE:

n   2:

n   3:

n   4:

3

 

 6

is in the IV quadrant.

11 11 24 13  12     2 2 2 2 13       2  is in the IV quadrant. NOTE: 3 6 6

 

11 11 36 25  18     2 2 2 2 25       4  NOTE: is in the IV quadrant. 3 6 6

 

11 11 48 37  24     2 2 2 2 37       6  is in the IV quadrant. NOTE: 3 6 6

 

: : : Answers:  

6.

7 11  6 n ;    6 n , where n is an integer 2 2

  tan  5    1 3  First, we must get the angle argument for the trigonometric function to be a single variable.

tan u  1 .

So, let u  5 



3

.

Thus, we obtain the equation

Now, determine where the solutions u will occur. Since tangent is positive in the I and III quadrants, the solutions for this equation occur in those quadrants. Find the reference angle u ' for the solutions u :

tan u  1 

tan u '  1 

u '  tan  1 1 

 4

The solutions in the I quadrant: The one solution u in the I quadrant, that is



between 0 and 2 , is u 

. Now, all the other solutions u in the I 4 quadrant are coterminal to this one solution. Thus, all the solutions u in the I

quadrant are given by u 



 2 n , where n is an integer. These are the

4

solutions for u. Now, find the solutions for  . Since u  5 

u 

 4

 2 n , then 5 

 3



 4

sides of the equation. Thus, 5 



 12

 4

 2 n . First, subtract  2n  

 3

=

 3

 3

and

from both

3 4  2n   = 12 12

 2 n . Now, divide both sides of this equation by 5 in order to solve

for  . Thus,   

 60



2 n 24 n    24 n   = =  = 5 60 60 60

( 24 n  1)  , where n is an integer. 60

Thus, some of the solutions in the I quadrant that are obtained by using ( 24 n  1)   , where n is an integer, are: 60

n  0:

( 0  1)      60 60 60 4 3  ...