2. Warm-up
How many possible solutions are there to
the following?
a.sin x = 2 b.cos y = 0
c. tan z = 3 d.sin a = β 1
2
3. Warm-up
How many possible solutions are there to
the following?
a.sin x = 2 b.cos y = 0
None
c. tan z = 3 d.sin a = β 1
2
4. Warm-up
How many possible solutions are there to
the following?
a.sin x = 2 b.cos y = 0
None Infinite
c. tan z = 3 d.sin a = β 1
2
5. Warm-up
How many possible solutions are there to
the following?
a.sin x = 2 b.cos y = 0
None Infinite
c. tan z = 3 d.sin a = β 1
2
Infinite
6. Warm-up
How many possible solutions are there to
the following?
a.sin x = 2 b.cos y = 0
None Infinite
c. tan z = 3 d.sin a = β 1
2
Infinite Infinite
15. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
Ο to the nearest thousandth.
16. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
Ο to the nearest thousandth.
cos x = .456
17. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
Ο to the nearest thousandth.
cos x = .456
cosβ1
(cos x ) β1
= cos (.456)
18. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
Ο to the nearest thousandth.
cos x = .456
cosβ1
(cos x ) β1
= cos (.456)
x β 1.097
21. Example 1
b. Estimate all solutions between 0 and 2Ο.
We already know one: 1.097
In which quadrant is this answer?
22. Example 1
b. Estimate all solutions between 0 and 2Ο.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I
23. Example 1
b. Estimate all solutions between 0 and 2Ο.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I
In which quadrant will our other answer be?
24. Example 1
b. Estimate all solutions between 0 and 2Ο.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I
In which quadrant will our other answer be?
Quadrant IV
25. Example 1
b. Estimate all solutions between 0 and 2Ο.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I
In which quadrant will our other answer be?
Quadrant IV
2Ο - 1.097 β
26. Example 1
b. Estimate all solutions between 0 and 2Ο.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I
In which quadrant will our other answer be?
Quadrant IV
2Ο - 1.097 β 5.186
28. Example 1
c. Describe all real solutions.
Now, weβre going to need all equivalent
values to 1.097 and 5.186. How do we
account for ALL of them?
29. Example 1
c. Describe all real solutions.
Now, weβre going to need all equivalent
values to 1.097 and 5.186. How do we
account for ALL of them?
1.097 + 2Οn
30. Example 1
c. Describe all real solutions.
Now, weβre going to need all equivalent
values to 1.097 and 5.186. How do we
account for ALL of them?
1.097 + 2Οn 5.186 + 2Οn
31. Example 1
c. Describe all real solutions.
Now, weβre going to need all equivalent
values to 1.097 and 5.186. How do we
account for ALL of them?
1.097 + 2Οn 5.186 + 2Οn
Here, n represents any integer value,
positive or negative
35. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x β 1 = 0
36. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x β 1 = 0
(2x β 1) (x + 1) = 0
37. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x β 1 = 0
(2x β 1) (x + 1) = 0
(2x β 1) = 0
38. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x β 1 = 0
(2x β 1) (x + 1) = 0
(2x β 1) = 0 ( x + 1) = 0
39. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x β 1 = 0
(2x β 1) (x + 1) = 0
(2x β 1) = 0 ( x + 1) = 0
x = 1
2
40. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x β 1 = 0
(2x β 1) (x + 1) = 0
(2x β 1) = 0 ( x + 1) = 0
x = 1
2 x = β1
42. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly!
43. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
44. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
Let u = tan x
45. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
Let u = tan x
2
Then 3u + 4u + 1 = 0
46. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
Let u = tan x
2
Then 3u + 4u + 1 = 0
(3u + 1) (u + 1) = 0
47. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
Let u = tan x
2
Then 3u + 4u + 1 = 0
(3u + 1) (u + 1) = 0
(3u + 1) = 0
48. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
Let u = tan x
2
Then 3u + 4u + 1 = 0
(3u + 1) (u + 1) = 0
(3u + 1) = 0 (u + 1) = 0
49. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
Let u = tan x
2
Then 3u + 4u + 1 = 0
(3u + 1) (u + 1) = 0
(3u + 1) = 0 (u + 1) = 0
u = β 1
3
50. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0
Ew! That looks ugly! Substitute!
Let u = tan x
2
Then 3u + 4u + 1 = 0
(3u + 1) (u + 1) = 0
(3u + 1) = 0 (u + 1) = 0
u = β 1
3 u = β1
55. Substitute back in:
u = β 1
3 u = β1
tan x = β 1
3
(
tanβ1 tan x ) ( )
= tanβ1 β 1
3
56. Substitute back in:
u = β 1
3 u = β1
tan x = β 1
3
(
tanβ1 tan x ) ( )
= tanβ1 β 1
3
x β β18.43494882Β°
57. Substitute back in:
u = β 1
3 u = β1
tan x = β 1
3
tan x = β1
(
tanβ1 tan x ) ( )
= tanβ1 β 1
3
x β β18.43494882Β°
58. Substitute back in:
u = β 1
3 u = β1
tan x = β 1
3
tan x = β1
(
tanβ1 tan x ) ( )
= tanβ1 β 1
3
tanβ1
(tan x ) β1
= tan ( β1)
x β β18.43494882Β°
59. Substitute back in:
u = β 1
3 u = β1
tan x = β 1
3
tan x = β1
(
tanβ1 tan x ) ( )
= tanβ1 β 1
3
tanβ1
(tan x ) β1
= tan ( β1)
x β β18.43494882Β° x β β45Β°
60. Substitute back in:
u = β 1
3 u = β1
tan x = β 1
3
tan x = β1
(
tanβ1 tan x ) ( )
= tanβ1 β 1
3
tanβ1
(tan x ) β1
= tan ( β1)
x β β18.43494882Β° x β β45Β°
General solutions:
61. Substitute back in:
u = β 1
3 u = β1
tan x = β 1
3
tan x = β1
(
tanβ1 tan x ) ( )
= tanβ1 β 1
3
tanβ1
(tan x ) β1
= tan ( β1)
x β β18.43494882Β° x β β45Β°
General solutions:
( )
x β β18.43 + 180n Β° or x β β45 + 180n Β° ( )