Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Key
Name:
Trigonometry - Beyond the Right Triangles
TRIGONOMETRIC EQUATIONS
Solve each equation, finding all solutions exactly in radians, without your calculator.
1.
2 sin x + 1 = 0
2.
4 cos2 x - 1 = 0
cos 2 x  14
2sin x  1
cos x   12
sin x   12
7
 2 k
6
11
or x 
 2 k ,
6
k
x
x
3.
tan2 x = 3

3
4.
2
 2 k ,
3
cos x  tan x = cos x
cos x tan x  cos x  0
cos x  tan x  1  0
cos x  0
 k ,
or tan x  1
 if cos x  0, tan x 

 is undef., so we  x   k , k 
4
 can't have cos x  0 
3=0
2 cos(3x) +
cos 3 x  
 2 k
k
k
5.
3
or x  
tan x   3
x

6.
3
2
5
3x  
 2 k
6
5 2k
x

,k
18
3
© 2005 Illinois Mathematics and Science Academy® Trig. 12.1
2 sin2 x = sin x
2sin 2 x  sin x  0
sin x  2sin x  1  0
sin x  0 or sin x 
x  k or x 
1
2

 2k
6
5
or x 
 2 k , k 
6
Rev. S06
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Key
Name:
Use identities to help solve the following. Give answers in radians, without your calculator.
7.
sin (2x) = sin x
8.
cos (2x) = 1 - sin x
1  2sin 2 x  1  sin x  0
2sin x cos x  sin x  0
sin x  2sin x  1  0
sin x  2 cos x  1  0
sin x  0 or cos x 
x  k or x  
9.

3
sin x  0 or sin x 
1
2
x  k or x 
10.
sin x  2 cos 2 x  1  0
sin x  0 or cos x  
x  k or x 

4


 2k
6
5
or x 
 2 k , k 
6
 2k , k 
sin (2x) = tan x
sin x
2sin x cos x 
cos x
2
2sin x cos x  sin x
1
2
2 cos x + 3 + sec x = 0
1
2 cos x  3 
0
cos x
2 cos 2 x  3cos x  1  0
 2 cos x  1 cos x  1  0
cos x   12
1
2
k
,k
2
x
or cos x  1
2
 2k or x    2k , k 
3
Solve each equation. Round answers to the nearest hundredth of a radian.
11.
3 sin x = cos x
12.
12 sin2 x + cos x - 11 = 0
12 1  cos 2 x   cos x  11  0
n.b. If cos x  0, then sin x  0,

so cos x cannot be 0 here.

3sin x cos x

cos x cos x
3 tan x  1
tan x  13
x  0.32  k , k 
© 2005 Illinois Mathematics and Science Academy® Trig. 12.2
12 cos 2 x  cos x  1  0
12 cos 2 x  cos x  1  0
 4 cos x  1 3cos x  1  0
cos x   14 or cos x  13
x  1.82  2k
or x  1.23  2k , k 
Rev. S06
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name:
13.
2 sec2 x + tan x - 5 = 0
2  tan 2 x  1  tan x  5  0
14.
Key
sin (2x) cos x + sin x cos (2x) = 0.2
sin  3x   0.2
2 tan 2 x  tan x  3  0
3x  sin 1  0.2   2k
 tan x  1 2 tan x  3  0
or 3x    sin 1  0.2   2k
tan x  1 or tan x   32
x  0.07 
x

4
2 k
3
2
or x  0.98 
,k
3
 k or x  0.98  k , k 
Beware! Not all equations have "nice" answers.
15.
Consider the equation cos x = x
a.
Graph y = cos x and y = x.
Determine the point of
intersection of these graphs.
b.
x  0.74
x  0.74
Consider the sequence:
k
if n  1

an  
cos(an 1 ) if n  1
Use recursion to find the fixed point.
Then compare your answer with the point
of intersection found in part a.
x  0.73908
Find all solutions such that -2 < x < 2. (Round answers to the nearest hundredth.)
16.
3 cos x = 2 + sin x
x  5.72,  1.21,
0.56, 5.07
17.
sin (3.4x) = sin x (Find some solutions.)
There are 13 intersection points in  2 , 2  . Using technology:
x  5.236, 4.998, 3.570, 2.618, 2.142, 0.714, 0
Alternately, consider that if sin A  sin B , then A  B  2k or A    B  2k , k  .
Here, we then have 3.4x  x  2k or 3.4x    x  2k
2.4x  2k or 4.4 x   2k  1 
x
 2k  1  , k 
2 k
or x 
2.4
4.4
© 2005 Illinois Mathematics and Science Academy® Trig. 12.3
Rev. S06