Download 5.3

JMerrill, 2010


It will be imperative that you know the
identities from Section 5.1. Concentrate on
the reciprocal, quotient, and Pythagorean
identities.
The Pythagorean identities are crucial!


Solve sin x = ½
Where on the circle does the sin x = ½ ?
Solve for [0,2π]
x 
 5
,
6 6
Particular
Solutions
Find all solutions
General
solutions
5
x   2n  ,
 2n 
6
6

Solve 2sin   1  0 for [0 ,360 )
2
2sin   1  0
2
2sin   1
2
1
sin  
2
2
2
sin   
2
o
o
There are 4 solutions
because sin is positive
in 2 quadrants and
negative in 2 quadrants.
  45o ,135o ,225o ,315o

Find all solutions to: sin x +
2
= -sin x
sinx  sinx  2  0
2sinx   2
sin x  
x
2
2
5
 2n
4
and
x
7
 2n
4

Solve 3tan2 x  1  0 for [0,2]
3
tan x  
3
 5 7  11
x , ,
,
6 6 6 6
Solve sin x tan x  3sin x for [0, 2 )
sin x tan x  3sin x  0
Round to nearest
hundredth
sin x(tan x  3)  0
sin x  0
or tan x  3
x  0,3.14
x  1.25,4.39

Solve cot x cos2 x  2cot x in [0,2)
These 2 solutions
are true because
of the interval
specified. If we
did not specify and
interval, you
answer would be
based on the
period of tan x
which is π and your
only answer would
be the first
answer.
cot x cos2 x  2cot x  0
cot x(cos x  2)  0
2
cot x  0
x
 3
,
2 2
cos2 x  2  0
cos2 x  2
cos x   2
DNE   (Does Not Exist)
No solution
Verify graphically

Quick review of Identities
Reciprocal Identities
sin  
1
csc
1
cos 
sec
1
tan  
cot 
Also true:
1
csc 
sin 
1
sec 
cos
1
cot  
tan 
Quotient Identities
sin 
tan  
cos
cos
cot  
sin 
Pythagorean Identities
sin   cos   1
2
2
1  cot   csc 
2
2
tan   1  sec 
2
2
These are crucial!
You MUST know
them.
sin2
cos2
1
tan2
sec2
cot2
csc2
(Add the top of the triangle to = the bottom)

Strategies
 Change all functions to sine and cosine (or at least
into the same function)
 Substitute using Pythagorean Identities
 Combine terms into a single fraction with a
common denominator
 Split up one term into 2 fractions
 Multiply by a trig expression equal to 1
 Factor out a common factor
x  3x  4  0
2
( x  1)( x  4)  0
( x  1)  0 or ( x  4)  0
x  1
x4
sin x  sin x  cos x
2
2
Hint: Make the words match
so use a Pythagorean identity
sin x  sin x  1  sin x
2
2
sin x  sin x  1  sin x  0
2
2
2sin x  sin x  1  0
2
Quadratic: Set = 0
Combine like terms
Factor—(same as 2x2-x-1)
(2sin x  1)(sin x  1)  0
1
sin x 
or sin x  1
2
x
 7 11
,
2 6
,
6
Solve 2sin  cos for [0 ,360 )
o
2sin  cos
cos
2
sin 
o
2  cot 
1
 tan 
2
  26.6 , 206.6
o
o

You cannot divide both sides by a common
factor, if the factor cancels out. You will lose
a root…
sin x tan x  3sin x
sin x tan x 3sin x

sin x
sin x
tan x  3
Common factor—
lost a root
2sin  cos
2
cos
sin 
2  cot 
1
 tan 
2
No common factor = OK

Sometimes, you must square both sides of an
equation to obtain a quadratic. However, you
must check your solutions. This method will
sometimes result in extraneous solutions.






Solve cos x + 1 = sin x in [0, 2π)
There is nothing you can do. So, square
both sides
(cos x + 1)2 = sin2x
cos2x + 2cosx + 1 = 1 – cos2x
2cos2x + 2cosx = 0
Now what?
Remember—you
want the words to
match so use a
Pythagorean
substitution!




2cos2x + 2cosx = 0
2cosx(cosx + 1) = 0
2cosx = 0
cosx + 1 = 0
cosx = 0
cosx = -1
x
 3
,
2 2
x 
cos
x
 3
,
2 2

 1  sin
2
0 1  1
x 

2
3
3
cos
 1  sin
2
2
0  1  1
cos   1  sin 
1  1  0






Solve 2cos3x – 1 = 0 for [0,2π)
2cos3x = 1
cos3x = ½
Hint: pretend the 3 is not there and solve
cosx = ½ .
Answer:
1 1
x  cos
2
But….
 5
x ,
3 3




In our problem 2cos3x – 1 = 0
What is the 2? amplitude
What is the 3? frequency
This graph is happening 3 times as often as
the original graph. Therefore, how many
answers should you have?
6
1
x  cos
2
 5
x ,
3 3
1
Add a whole circle to each of these
7  11
,
3 3
And add the circle
once again.
13 17 
,
3
3
Final step: Remember we pretended the 3 wasn’t there,
but since it is there, x is really 3x:
3x 
 5 7 11 13 17
3
,
3
,
3
,
,
,
3
3
3
 5 7 11 13 17
So, x  , ,
,
,
,
9 9 9 9
9
9

Work the problems by yourself. Then
compare answers with someone sitting next
to you.
Round answers:
o
o
191.5
,348.5
1. csc x = -5 (degrees)

2. 2 tanx + 3 = 0 (radians)

3. 2sec2x + tanx = 5 (radians)


2.16, 5.30
2.16, 5.30, .79, 3.93

4. 3sinx – 2 = 5sinx – 1
7  11
,
6 6

5. cos x tan x = cos x
 3  5
, , ,
2 2 4 4

6. cos2  - 3 sin  = 3
3
2