Download Trig 11.2 Solving Equations key - IMSA

Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
Key
Trigonometry: Modeling the Seas
SOLVING TRIGONOMETRIC EQUATIONS
For these problems, it will be helpful to recall your work on maximum and minimum points.
1.
2.
3.
Consider: sin(x) = 1.

a.
State one solution:
b.
How many solutions are there? How far apart are consecutive solutions?
Infinitely many, 2 (one period) apart.
c.
Combine this information to find the location of all solutions to this equation.
(Use k in your answer, and state the possibilities for it.)

x   2 k , k 
2
Consider: sin(2x) = 1.
2

a.
State one solution:
b.
How many solutions are there? How far apart are consecutive solutions?
Infinitely many,  (one period) apart.
c.
Combine this information to find the location of all solutions to this equation.

x   k , k 
4
4
Solve each of the following by finding a "starting" point and determining the distance
between consecutive values. (Note that "solve" requires finding all real solutions.)
a.
sin(x/3) = 1
b.
sin(2x) = 0
period  6
x
period  
3
 6 k , k 
2
© 2005 Illinois Mathematics and Science Academy®
x
Trig. 11.1
k
,k
2
Rev. S05
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
3.
Continued.
c.
sin(3x) = –1
d.
3x  32  2k
2k
x 
,k
2
3
4.
a.
sin(x/2) = 0
x
2

Key
 k
x  2 k , k 
Solve: cos(x) = 1. (Use the same process as above.)
x  2 k , k 
b.
Solve: cos(3x) = 1.
3 x  2 k
x
5.
2 k
,k
3
Solve.
a.
cos(x/3) = 1
x
3
b.
 2k
x  6 k , k 
cos(6x) = –1
c.
x
d.
6 x    2 k
x
6.
6

4

k
,k
2
 2x 
cos    0
 3
2x
3
k
,k
3

 2  k
x
3 3k

,k
4
2
Solve: tan(x) = 1
x
7.

cos(2x) = 0
2 x  2  k

4
 k , k 
Solve.
a.
tan(3x) = 1
3x  4  k
b.
x
4

k
x 
,k
12 3
© 2005 Illinois Mathematics and Science Academy®
tan(x/4) = 0
 k
x  4 k , k 
Trig. 11.2
Rev. S05
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
7.
Continued.
tan(2x)  3
c.
2 x  3  k
x
8.


6
k
,k
2
Consider sin(x) 
a.
d.
Key
tan(x/5) + 1 = 0
x
3
5  4  k
x
15
 5k , k 
4
1
.
2
Within the period [0, 2), how many solutions are there? State them:

5
6
6
2 solutions.
b.
State all solutions over the real numbers. (Note that a complete solution will
involve two "pieces.")
x
9.

5
 2k , k 
6
 2k or x 
6
Solve.
a.
sin(x) 
x

3
2
b.
 2 k
x
3
2
or x 
 2 k , k 
3
10.
sin(x) 
or x 
 2
2

4
 2 k
5
 2 k , k 
4
1
. We'll take a slightly different approach for this equation.
2
How does 2x relate to your solutions to problem 8?
Consider sin(2x) 
a.
2x 
b.

6
 2k or 2x 
5
 2 k , k 
6
Now solve for x. (Check your answer for reasonableness. How far apart should
the answers be in each group?)   apart (the period for y  sin  2 x  )
x

12
 k or x 
5
 k , k 
12
© 2005 Illinois Mathematics and Science Academy®
Trig. 11.3
Rev. S05
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
11.
Solve these problems with a phase shift.
  3

a.
sin  x   

6
2
x  6   3  2k

x
6
x  6 
or
 2 k
x
4
3
4  1

cos  x 


3  2
x  43   3  2k
b.
 2 k
3
 2 k
2
Key
x    2k
5
 2 k
3
k
k
12.
x
or
Solve.
 x  1
sin   
 2 2
x

x
7
2   6  2 k
2  6  2 k
or

7
x    4 k
x
 4 k
3
3
k
a.
sin 4 x  
c.
4 x  3  2k
x

12

k
2
3
2
4x 
or
x
 2 k
2
3


6
k
2
k
1
2
3x   
sin 3x    
e.
3x    6  2k
3x 
x
7
6
 2 k
3x 
or
7 2k

18
3
x
4
3
11
6
2
2
cos 3x  
b.
3x   4  2k
x

12

2 k
,k
3
 2
2

2 x   4  2 k
2 x  54  2k
or

5
x    k
x
 k
8
8
k
d.
sin 2x  
f.
3
 3x 
cos   
 2
2
 2 k
 2 k
3x
2
11 2k

18
3
  6  2k
x

9

4 k
,k
3
k
 x  5   3
cos 

 4 
2
x 5
5
4   6  2 k
g.
cos  3x    12
x  5   103  8k
25
x
 8k
3
or
x
3
5
x
 8k
3
k
 n.b. x  3  8k ~ x  253  8k as

 x
2 cos    1  0
 3
h.
24
3
  23  2k
x  2  6k , k 
 8 
© 2005 Illinois Mathematics and Science Academy®
Trig. 11.4
Rev. S05