Download A simple equation such as 2x + 2 = 3 has a simple solution, x =

Handout for Section 5.3
Solving Trigonometric Equations
1
. With a Trigonometric
2
Equation such as 2cos  2  3 solving for  is somewhat more complicated.
Let’s take a look at what we can do with 2cos  2  3
A simple equation such as 2 x  2  3 has a simple solution, x 
First let’s solve for cos  .
2 cos  1
cos 
1
2
If we translated the last line into English we would say, “What value(s) of  will make cosine
1
equal to ?”
2
Now to solve for  let’s draw ourselves a picture while thinking about what angles will make
1
cosine equal to  ?
2

is probably the first angle that comes to mind. So there you have it! For the equation
3

2cos  2  3 , the solution is 60 or .
3
60 or
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1
However just like if we had the equation x2  4 there is more than one solution.
x2  4
x2   4
x  2
Both 2 and 2 will work. So we must always consider every possible answer for an equation.
Let’s go back to the equation 2cos  2  3 and see if any other angles other than 60 make
1
5
cosine equal to  . The next most obvious angle is probably 300 or
which is in quadrant
2
3
IV.
So far we have two angles that satisfy the equation, however that’s not all. Add a complete
revolution to the angle 60 and the angle 420 is coterminal with 60 . Also 780 ,1140 ,1500
etc… Likewise the angle 660 is coterminal with 300 ,1020 ,1380 ,1740 etc… So in reality
we have an infinite number of solutions to this equation. Since it is tedious to write them all out,
we find a pattern for these values. If we keep adding 360 to both 60 and 300 this will give us
all the possible solutions for the equation.

7 13
(The same is true when working with radians.
is coterminal with
,
, etc… Just like
3
3
3
adding 360 before except now we are adding 2 which is one complete revolution)
So you write your answer like this:
  60  k  360 ,   300  k  360
or in radians:
5
 2k
3
3
Where k is any given integer (including negatives)


 2k ,  
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2
(Note that one may generate a table of  ' s for various values of k in this way.)

k

3
 2k


 2k
3
k
  60  k  360
  300  k  360
2
660
420
1
300
60
11
3

5
3

0

3
5
3
0
60
300
1
7
3
11
3
1
420
660
2
13
3
17
3
2
780
1020
2

1

etc…
7
3

3
etc…
The answer we have given is for ALL possible values. Occasionally we may not need all the
values, only those greater than or equal to 0 and less than 360 (or in radians, 0    2 ). This
set of values is called the set of fundamental values, those values of  where 0    360 (or
0    2 ). So the fundamental values for the previous equation would be:
  60 , 300
or in radians:


3
,
5
3
Let’s try another.
csc  2  0
First we solve for csc .
csc   2
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3
Now we could think of an angle  that makes cosecant equal to  2 , but maybe there is an
1
easier way. As you should know csc 
. So we can manipulate the equation like so…
sin 
1
 2
sin 
Thus…
sin   
1
2
This might be easier to think about because some of us may be more familiar with sine than
1
cosecant. So now let’s think of some angles that make sine equal to 
. First recall since sine
2
is negative,  must be in quadrant III or IV.
and
The two angles,   225 and 315 , are the fundamental solutions for this equation. To find all
the solutions remember we add revolutions, ( 360 , 2 ) to each of them infinitely. All solutions
would be written:
  225  k  360 ,   315  k  360
or in radians:

5
7
 2k ,  
 2k
4
4
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4
Next we’ll try:
3 tan   1  0
First we solve for tan  .
3 tan   1
1
tan  
3
What values of  make a tangent equal to
1
3
? Now we think of an angle that makes this
statement true.
and
(Note: the angle must be in quadrant I or III so that tangent will be positive.)
So our fundamental solutions are:
  30 , 210 ( 

6
,
7
)
6
For all solutions, both tangent and cotangent have a different period than the other trig functions.
In our previous problems we had answers involving k  360 . With tangent and cotangent the
two solutions are directly across from each other on the circle (i.e. 180 opposite each other).
Thus we only need add 180 to generate all the solutions.
So the answer for all solutions is:
  30  k 180
or in radians:


6
 k
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5
Trigonometric Equations With Exponents
Ok let’s move on and do one involving exponents.
4 sin 2   1  0
4 sin 2   1
sin 2  
1
4
Now we take the square root of both sides to solve for sin  .
sin 2   
1
4
Remember if we take the square root of both sides the right side has to be  .
sin   
1
2
So essentially we have:
1
1
sin   , 
2
2
First find angles that make sine equal to
1
.
2
and
1
Now let’s find the angles that make sine equal to  .
2
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and
So we now know that our fundamental solutions are:
  30 , 150 , 210 , 330
And in radians:


6
,
5 7 11
,
,
6
6
6
You may notice that 30 and 210 are supplementary to (directly opposite of) each other.
Likewise with 150 and 330 . This means that we will only need two equations for all solutions,
both involving k 180 .
All solutions:
  30  k  180 ,   150  k  180
In radians:


6
 k ,  
5
 k
6
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Simplification of Trigonometric Equations
Sometimes you have to simplify a problem before it can be solved.
cos 2  tan 
 1
sin  (cos 2  sin   sin 3  )
First factor out sin  on the bottom.
cos 2  tan 
 1
sin 2  (cos 2   sin 2  )
Note: cos2   sin 2   1
cos 2 
tan   1
sin 2 
cos 2  sin 
 1
sin 2  cos 
cos 
 1
sin 
Note:
cos 
 cot 
sin 
cot   1
Note:
1
 tan 
cot 
1
1

cot  1
tan   1
All Solutions:
  135  k  180
3

 k
4
Fundamental Solutions:
  135 ,315
3 7

,
4
4
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8
Products of Trigonometric Equations
Sometimes you will encounter a product of trigonometric functions like so.
( 3 cot   1)(4 cos 2   3)  0
You may recall solving quadratics with similar form such as (2x  1)( x  9)  0 . When we have
a product we set each factor equal to zero and solve. (Because if (a)(b)  0, then either a  0 or
b  0.)
Case I:
Case II:
2x 1  0
x 9  0
2 x  1
x9
1
x
2
1
or 9 .
2
We do the same for our trigonometric product. Let’s solve both factors for  .
So our answers are x 
Case I:
Case II:
3 cot   1  0
3 cot   1
cot  
1
4 cos 2   3  0
4 cos 2   3
cos 2  
3
3
4
3
2
tan    3
cos   
All Solutions:
All Solutions:
  60  k 180

   k
  30  k 180 , 150  k 180

5
   k ,
 k
Fundamental Solutions:
Fundamental Solutions:
  60 , 240
 4
 ,
  30 ,150 , 210 ,330
 5 7 11
 , , ,
3
3
3
6
6
6
6
6
6
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The final solution is the union of the two solution sets above.
All Solutions:
  30  k 180 , 60  k 180 , 150  k 180


5
   k ,  k ,
 k
6
3
6
Fundamental Solutions:
  30 , 60 ,150 , 210 , 240 ,330
  5 7 4 11
 , , , , ,
6 3
6
6
3
6
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10
Trigonometric Equations With a Quadratic Form
You might also come across an equation with a somewhat quadratic form.
2sin 2   sin   1  0
The factored version might not be easy to see when written like this. However if we substitute
a  sin  then we can change it to.
2a2  a 1  0
Now factor it.
(2a  1)(a  1)  0
Now going back to the original variable, substitute a  sin  .
(2sin 1)(sin 1)  0
And solve the equation for
2sin   1  0
1
sin  
2
sin   1  0
or
sin   1
All Solutions:
  90  k  360 , 210  k  360 , 330  k  360

7
11
   2k ,
 2k ,
 2k
2
6
6
Fundamental Solutions:
  90 , 210 , 330
 7 11
 ,
,
2
6
6
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11
Trigonometric Equations With Decimal Answers
Unfortunately all trigonometric equations don’t have nice answers. Let’s take the following
example.
3sec  5  0
Start it like a normal equation.
3sec   5
5
3
3
cos  
5
sec  
The triangles that satisfy this condition would like look this.
 3
You’ll have to use your calculator to find this angle. To do so put cos 1   in your calculator.
5
Provided you are in degree mode you should obtain 53.13 as your answer. ( .9273 if you are in
radian mode.) This is  . We can see from the picture however that this is not the only answer.
By observation we can see that  and  are opposites of each other. Or in other words:    
So we would have 53.13 ( .9273 ) and 53.13 ( .9273 ) as our answers. However we want our
answers to be between 0 and 360 including 0 but not 360 . To obtain the angle coterminal
with 53.13 ( .9273 ) between 0 and 360 we add a complete revolution or 360 (2 ) . Doing
so we obtain 306.87 or 5.3559 . You can test your answers by taking the cosine of both
answers. If they are equal you know your second angle is correct provided your first angle is
too.
(Note: You may have to go up to at least 8 decimal points of accuracy for your answers to come
out the same. These previous answers are rounded off to two decimal points in degrees and four
decimal points in radians.)
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All Solutions:
  53.13  k  360 , 306.86  k  360
  .9273  2k , 5.3559  2k
Fundamental Solutions:
  53.13 ,306.86
  .9273, 5.3559
Now don’t get scared if that was a little confusing. Or maybe a lot confusing. We’ll try another
one and hopefully it’ll start to make sense. Let’s try
sin  
5
13
 5
Use your calculator to take sin 1   . You will get 22.62 or .3948 in radians. So now we
 13 
draw this angle and any other angles that might have an equivalent sine value.
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So our first angle   22.62 and now we need to find what  is. If we look closely we can see
that  and  are equal. Also note that     180 .
Now we already know that   22.62 so put it into the equation and solve for  .
22.62    180
  157.38
And now we have our solutions.
All Solutions:
  22.62  k  350 ,157.38  k  360
  .3948  2k , 2.7468  2k
Fundamental Solutions:
  22.62 ,157.38
  .3948, 2.7468
Before continuing on to any new concepts, let’s combine the ones we’ve learned so far to solve
this problem:
6sin 2 x  cos x  5  0
There is a problem. It won’t solve like a normal quadratic because we have a mixture of sin and
cosine. We need to change it so that either everything is cosine or everything is sine. Recall:
sin2 x  1  cos2 x
Using this we can change everything to cosine.
6(1  cos 2 x)  cos x  5  0
6  6 cos 2 x  cos x  5  0
6 cos  x  cos x  1  0
Now we are able to factor this like the other quadratics.
(3cos x 1)(2cos x 1)  0
Now solve each one for cosx.
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14
3cos x  1  0
cos x 
2 cos x  1  0
or
1
3
cos x  
1
2
Solving the first we obtain:
x  70.53  k  360 , 289.47  k  360
x  1.2310  2k , 5.0522  2k
And the second:
x  120  k  360 , 240  k  360
x
2
4
 2k ,
 2k
3
3
So…
All Solutions:
x  70.53  k  360 , 120  k  360 , 240  k  360 , 289.47  k  360
x  1.2310  2k ,
2
4
 2 k ,
 2k , 5.0522  2k
3
3
Fundamental Solutions:
x  70.53 , 120 , 240 , 289.47
x  1.2310,
2 4
,
, 5.0522
3
3
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15
Multiple Angle Variables
Finally we’ll take a look at equations that have an argument other than a single variable. Let’s
take cos(4  20 )  1
We begin by substituting a  4  20
So now we have cos a  1
Not too hard. cos a  1 at   180 ( in radians) .
So all solutions would be:
  180  k  360
    2k
However  was not our original variable so we certainly can’t end with it. We need to go back
and substitute a  4  20 into our “all solutions”.
4  20  180  k  360
Now solve for  .
4  20  180  k  360
4  160  k  360
4 160
360

k
4
4
4
  40  k  90
Now we have the real solutions.
All Solutions:
  40  k  90
2


k
9
4
Finding the fundamental solutions for an equation like this is a little more involved. To find the
values between 0 and 360 we plug in values for k .
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16
  40  k  90
k

1
0
50
1
2
3
4
40
130
220
310
400
Now that we have gone below 0 and above 360 we take the values that fall between them as
our fundamental values.
Fundamental Solutions:
  40 ,130 , 220 ,310
2 13 11 31

,
,
,
9
NOTE:
18
9
18
If you have a TI-82, TI-83, or TI-83+ read pages 23-27 in your graphing
calculator manual.
If you have a TI-86 read pages 131-135 in your graphing calculator manual.
You will need to know how to solve equations using your graphing calculator.
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17