1. The document discusses solving trigonometric equations and finding their general solutions. It provides examples of solving equations using inverse trig functions, factoring, and substitution. 2. General solutions to trig equations involve adding integer multiples of the period (2π or 180°) to the solutions to account for all possibilities in the entire domain. 3. Examples show solving equations like cosx = 0.456 by taking the inverse cosine and factoring equations like 3tan^2x + 4tanx + 1 = 0 to find specific solutions and the general form.

1. Notes 5-7
General Solutions to Trigonometric
Equations

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 = − 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 = − 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 = − 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 = − 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 = − 1
2
Infinite Infinite

7. Trigonometric Equation:

8. Trigonometric Equation:
An equation where the variable is within
one of the trig functions

9. 3 types of domains

10. 3 types of domains
Restricted domains that allow us to
find the inverse functions

11. 3 types of domains
Restricted domains that allow us to
find the inverse functions
One period

12. 3 types of domains
Restricted domains that allow us to
find the inverse functions
One period
All real numbers

13. General Solution:

14. General Solution:
The solution that takes into account all
possible solutions for a trig equation

15. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
π to the nearest thousandth.

16. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
π to the nearest thousandth.
cos x = .456

17. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
π to the nearest thousandth.
cos x = .456
cos−1
(cos x ) −1
= cos (.456)

18. Example 1
Consider the equation:
cos x = .456
a. Estimate the solution between 0 and
π to the nearest thousandth.
cos x = .456
cos−1
(cos x ) −1
= cos (.456)
x ≈ 1.097

19. Example 1
b. Estimate all solutions between 0 and 2π.

20. Example 1
b. Estimate all solutions between 0 and 2π.
We already know one: 1.097

21. Example 1
b. Estimate all solutions between 0 and 2π.
We already know one: 1.097
In which quadrant is this answer?

22. Example 1
b. Estimate all solutions between 0 and 2π.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I

23. Example 1
b. Estimate all solutions between 0 and 2π.
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 2π.
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 2π.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I
In which quadrant will our other answer be?
Quadrant IV
2π - 1.097 ≈

26. Example 1
b. Estimate all solutions between 0 and 2π.
We already know one: 1.097
In which quadrant is this answer?
Quadrant I
In which quadrant will our other answer be?
Quadrant IV
2π - 1.097 ≈ 5.186

27. Example 1
c. Describe all real solutions.

28. Example 1
c. Describe all real solutions.
Now, we’re 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, we’re going to need all equivalent
values to 1.097 and 5.186. How do we
account for ALL of them?
1.097 + 2πn

30. Example 1
c. Describe all real solutions.
Now, we’re going to need all equivalent
values to 1.097 and 5.186. How do we
account for ALL of them?
1.097 + 2πn 5.186 + 2πn

31. Example 1
c. Describe all real solutions.
Now, we’re going to need all equivalent
values to 1.097 and 5.186. How do we
account for ALL of them?
1.097 + 2πn 5.186 + 2πn
Here, n represents any integer value,
positive or negative

32. Factoring!

33. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!

34. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:

35. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x − 1 = 0

36. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x − 1 = 0
(2x − 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 − 1 = 0
(2x − 1) (x + 1) = 0
(2x − 1) = 0

38. Factoring!
Remember waaaaaaaaaay back to Advanced
Algebra when you did factoring? Because
you should!
Factor the following:
2
2x + x − 1 = 0
(2x − 1) (x + 1) = 0
(2x − 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 − 1 = 0
(2x − 1) (x + 1) = 0
(2x − 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 − 1 = 0
(2x − 1) (x + 1) = 0
(2x − 1) = 0 ( x + 1) = 0
x = 1
2 x = −1

41. Example 2
Solve for x in degrees:
2
3 tan x + 4 tan x + 1 = 0

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 = − 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 = − 1
3 u = −1

51. Substitute back in:

52. Substitute back in:
u = − 1
3

53. Substitute back in:
u = − 1
3 u = −1

54. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3

55. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3
(
tan−1 tan x ) ( )
= tan−1 − 1
3

56. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3
(
tan−1 tan x ) ( )
= tan−1 − 1
3
x ≈ −18.43494882°

57. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3
tan x = −1
(
tan−1 tan x ) ( )
= tan−1 − 1
3
x ≈ −18.43494882°

58. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3
tan x = −1
(
tan−1 tan x ) ( )
= tan−1 − 1
3
tan−1
(tan x ) −1
= tan ( −1)
x ≈ −18.43494882°

59. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3
tan x = −1
(
tan−1 tan x ) ( )
= tan−1 − 1
3
tan−1
(tan x ) −1
= tan ( −1)
x ≈ −18.43494882° x ≈ −45°

60. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3
tan x = −1
(
tan−1 tan x ) ( )
= tan−1 − 1
3
tan−1
(tan x ) −1
= tan ( −1)
x ≈ −18.43494882° x ≈ −45°
General solutions:

61. Substitute back in:
u = − 1
3 u = −1
tan x = − 1
3
tan x = −1
(
tan−1 tan x ) ( )
= tan−1 − 1
3
tan−1
(tan x ) −1
= tan ( −1)
x ≈ −18.43494882° x ≈ −45°
General solutions:
( )
x ≈ −18.43 + 180n ° or x ≈ −45 + 180n ° ( )

62. Homework

63. Homework
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