Download Solutions to Math 112 Assignment #6: Trigonometric Equations - Prof. Jonathan Comes and more Assignments Mathematics in PDF only on Docsity!
Math 112 Jonny Comes Fall 2005 Assignment # Partial Solutions
From the Textbook:
- Section 7.5:
- Find all solutions in the interval [0, 2 π).
cot x cos x = cos x.
Solution:
cot x cos x = cos x ⇒ cot x cos x − cos x = 0
⇒ cos x(cot x − 1) = 0 ⇒ Case 1 : cos x = 0 or Case 2 : cot x − 1 = 0 Case 1: cos x = 0 implies x = ± arccos(0) + 2πk for any integer k. Since arccos(0) = π 2 this gives x = ± π 2 + 2πk for any integer k. Of these solutions the ones between 0 and 2π are
x = π 2 and x = − π 2 + 2π =^32 π.
Case 2:
cot x − 1 = 0 ⇒ cot x = 1 ⇒ tan x = 1 cot x
=^1
Now tan x = 1 implies x = arctan(1)+πk for any integer k. Since arctan(1) = π 4 this gives x = π 4 + πk for any integer k. Of these solutions the ones between 0 and 2π are
x = π 4
and x = π 4
So the final solution is
x =
π 2 or^ x^ =
3 π 2 or^ x^ =^
π 4 or^ x^ =
5 π
Find all solutions in the interval [0, 2 π).
tan x cos x = cos x.
Solution:
tan x cos x = cos x ⇒ tan x cos x − cos x = 0
⇒ cos x(tan x − 1) = 0 ⇒ Case 1 : cos x = 0 or Case 2 : tan x − 1 = 0 Case 1: cos x = 0 implies x = ± arccos(0) + 2πk for any integer k. Since arccos(0) = π 2 this gives x = ± π 2 + 2πk for any integer k. Of these solutions the ones between 0 and 2π are
x = π 2
and x = − π 2
Case 2: tan x − 1 = 0 ⇒ tan x = 1. Now tan x = 1 implies x = arctan(1)+πk for any integer k. Since arctan(1) = π 4 this gives x = π 4 + πk for any integer k. Of these solutions the ones between 0 and 2π are
x = π 4
and x = π 4
So the final solution is
x =
π 2 or^ x^ =
3 π 2 or^ x^ =^
π 4 or^ x^ =
5 π
Find all solutions in the interval [0, 2 π).
cos x csc x = 2 cos x.
Solution:
cos x csc x = 2 cos x ⇒ cos x csc x − 2 cos x = 0
⇒ cos x(csc x − 2) = 0 ⇒ Case 1 : cos x = 0 or Case 2 : csc x − 2 = 0
Now cos x = − 13 implies x = ± arccos
- 2πk for any integer k. Of these solutions the ones between 0 and 2π are
x = arccos
and x = − arccos
So the final solution is
x = 0 or x = π or x = arccos
or x = − arccos
+2π.
Additional Exercises: (Be sure to justify all your answers)
- True or False:
(a) If c is some real number, then we know the equation tan x = c has infinitely many solutions. Solution: TRUE. For ANY real number c we know tan x = c ⇒ x = arctan(c) + πk for any integer k. Thus there are infinitely many solutions, one for each integer. (b) If c is some real number, then we know the equation sin x = c has infinitely many solutions. Solution: FALSE. For example, set c = 2. We know sin x = 2 has zero solutions. [Any c-value greater then 1 or less than −1 will give a counterex- ample.] (c) If c is some real number, then we know the equation sin x = c has either zero or infinitely many solutions. Solution: TRUE. For any real number c between −1 and 1 we know
sin x = c =⇒
x = arcsin(c) + 2πk or x = π − arcsin(c) + 2πk for any integer k. Thus there are infinitely many solutions when − 1 ≤ c ≤ 1, one for each integer. On the other hand, when c is greater than 1 or less than −1 the equation sin x = c has no solution. So there is either zero or infinitely many solutions.
(d) If sec x = c where c > 1, then we know
x =
arccos(c) + 2πk^ or^ x^ =^ −^
arccos(c) + 2πk
Solution: FALSE. For example if c = 2, then sec x = 2 implies cos x = 1/2, thus x = ± π 3 + 2πk
for any integer k. On the other hand, arccos(2) is undefined, so ± (^) arccos(^1 c) + 2πk is undefined for all integers k. (e) If sec x = c where c > 1, then we know
x = arccos
c
c
Solution: TRUE. sec x = c implies cos x = (^1) c which implies
x = 1 arccos(c)
2πk or x = − 1 arccos(c)
2πk
for any integer k.
