Transcript faculty.camosun.ca

Solving Trig Equations
Finding the angle when the trig function value is known.
I. Isolate the trig function: sin   0.23, cos   0.567,
tan   2.367, etc.
II. Check the domain: 00    3600 or 0    2 , is the
usual case for now. Note: this means you will probably have
two answers, since each tirg function is  in two quads and
 in two quads.
III. Identify which two quadrants the terminal side lies in.
IV. Find the reference angle, it is always positive:
00  ref angle  900
V. Calculate the measure of the angle in standard position!
NOTE: we can use all the algebraic techniques to solve trig
equations that we use to solve algebraic equations.
For example:
1. If degree 2, make one side = zero and factor,
2. If you can't factor then use the quad formula,
3. If degree  3, then you should be able to factor, or
possibly use the polynomial techniques to solve degree 3
or higher equations, ie. rational zeros theorem
In all of these cases, remember that when you get to isolating
the trig function: sin   0.567, or cos   0.126, or
tan  5.483, there are usually two answers for each value!
If we are solving over the set of real numbers, then there
are an infinite number of solutions (coterminal angles).
Solve the trigonometric equation: 2cos  x   2 3  3, over:
a) 0  x  2 ,
b)
 ,   , over the reals,
find exact values!
Step I. Isolate the trig function.
3
2cos  x    3,

cos  x   
, x lies in quads II & III,
2
 3 
1
0
Step II. Find the reference angle: ref   cos 

or
30

 2  6
Step III. Calculate the angle in standard position:
Sketch the angles:
Indicate the angles
in standard position:
Now we can calculate the angle
in standard position part a:
 5
Quadrant II:     
6
6
 7
Quadrant III:     
6
6
7
6

6

6
5 7
Therefore the solutions are:  
,
6 6
5
6
Part b: over the set of real numbers.
5
7
This means any angles coterminal with
&
,
6
6
since the period of the cosine function is 2
5
7
Solution is:
 2n ,
 2n , n  Integers
6
6
Solve the trigonometric equation: 2sin 2  x   1  2, over:
a) 0  x  2 ,
b)
 ,   , over the reals,
1
2sin  x   1,

sin  x   ,
2
x lies in quads I,II, III & IV,
2
2
 1  
reference   sin 
 4
 2
1
find exact values!

1
sin  x   
2
 45 
0
Sketch the reference angles in all four quadrants:
Now we can calculate the angles
in standard position:
Quad I ref    in standard position 
3
4
4
 5
 
4
4
Quad II  
Quad III
Quad IV


4
3
4


7
2  
4
4

4
5
4
7
4
 3 5
7
Therefore the solutions are: , , , &
4 4 4
4
Solve the trigonometric equation: 3tan 2  x   2 tan  x   3, over:
a) 0  x  2 ,
b)
 ,   , over the reals, answers accurate to
four decimals!
Since degree two, write in standard form and see if we can factor.
3tan 2  x   2 tan  x   3  0, can't factor, so use the quad formula,
a  3, b  2, c  3
b  b  4ac
tan  x  

2a
2
  2  
 2   4  3 3
2
2  3
2  4  36

6
2  40
2  2 10 1  10
tan  x  


6
6
3
Since tan  x will be + &  , we can expect four answers from 0  x  2
Now find the reference angle!
 1  10 


ref   tan 

0.624522886

1 1  10

ref   tan 
  0.94627344
3


 3 
1
Sketch the reference angles in all four quadrants:
In quad I, ref    in standard position,
x 0.9463
In quad II, x    0.624522886
x
2.5171
In quad III, x    0.94627344
x
4.0879
In quad IV, x  2  0.624522886
x 5.6587
Therefore the solutions for
part a) x 0.9463, 2.5171,4.0879, 5.6587
Therefore the solutions for
part a) x
0.9463, 2.5171, 4.0879, 5.6587
part b) x  0.9463  n 
x  2.5171  n 
 x  0.9463  n 
 or 
 n  Integers
x  4.0879  n 
 x  2.5171  n 
x  5.6587  n 
The shorter version at the right is because the period
of the tangent function is  radians or 180
0
All the positive answers differ by  radians, and all
the negative answers differ by  radians!
Solving trig equations involving a phase shift and/or multiple angles!
Solve the following trig equation over: a) 0    2 , b)  , .


2sin      3  0
6

Step I: isolate the trig function.
  3

sin    
6
2


 3
Step II: let x    , so we have: sin  x  
6
2
Step III: Find the reference angle,
Angle x is in quadrants III & IV (since negative),
 3 
reference   sin 
 
 2  3
1
Sketch the reference angles in quad III & IV:

4
Quad III angle:   
3
3
 5
Quad IV angle: 2  
3
3
Now these two angles have

been shifted
left.
6
 4 5
So x    
,
6
3 3
4  5 
 
 ,

3 6 3 6
7 9  3 

,  
6 6  2 
4
3
5
3
Therefore the solutions for 0    2 are:
7 3

,
6 2
Therefore the solutions for
 ,  
are:
7
3

 2n ,
 2n , n  Integers
6
2
Solve a trig function involving multiple or half angles:
Solve the following trig equation over: a) 0    2 , b)  , .
3 tan  2 x   1  0
1
Step I: isolate the trig function: tan  2 x  
3
1
Step II : let   2 x

tan   
3
 is in quads II & IV, now find the reference angle:
 1  
Step III : ref   tan 
 6
 3
1
Sketch the reference angles in quad II & IV:

5
Quad II angle :   
6
6
 11
Quad IV angle : 2  
6
6
5 11 17 23
So now we have :   2 x 
,
,
,
6 6
6
6
5 11 17 23
 x
,
,
,
12 12 12 12
5
6
11
6
Over the set of real numbers: since period is
5 n
solution is:

, n  Integers
12 2
This is the graph of y  tan  2x 
1
For y  
3

2
,