Question

when is 2cos x + 1 >0 , < 0 AND = 0

Answer 1

2cos x + 1 > 0
2cos x > -1
cos x > -1/2
x > arccos -1/2

2cos x + 1 < 0
2cos x < -1
cos x < -1/2
x < arccos -1/2

2cos x + 1 = 0
2cos x = -1
cos x = -1/2
x = arccos -1/2

Answer 2

yes, all the solutions please

Answer 3

arccos (-1/2) = 120 degrees, 240 degrees, (120+360) degrees, (240+360) degrees, (120+360+360) degrees, etc.

Answer 4

increasing when 2pi*n < x < 2pi/3 + 2pi (n+1) , or 4pi/3 +2pi*n < x < 2pi + 2pi(n+1)

Answer 5

solving when 2cos x + 1 > 0

Answer 6

what is the entire question?

Answer 7

A) Domain, B)Intercepts, C) symmetry, D)Asymptotes, E) Intervals, F) Intervals of Increase or Decrease, G) Concavity and Points of Inflection, H) Sketch the Curve y = sinx/(2 + cosx)

Answer 8

what? that isn't a question.

Answer 9

y = sinx/(2 + cosx
find A) Domain, B)Intercepts, C) symmetry, D)Asymptotes, E) local max or min F) Intervals of Increase or Decrease, G) Concavity and Points of Inflection, H) Sketch the Curve

Answer 10

ah, okay. Which one can't you do?

Answer 11

all of em

Answer 12

Well, I can't do some of them either. This is a different question from the first right?

Answer 13

lets do intervals of increase and decrease

Answer 14

here are the derivatives
y = sinx/(2 + cosx)
y ' = [ 2 cos (x) + 1 ] / ( cos x + 2)^2
y '' = [ 2 sin x ( cos x - 1) ] / ( cos (x) + 2) ^ 3

Answer 15

http://www.youtube.com/watch?v=-W4d0qFzyQY

Answer 16

increasing on (2pi*k, 2pi/3 + 2pi (k+1) union (4pi/3 +2pi*k, 2pi + 2pi(k+1) )

Answer 17

do you agree?

Answer 18

we know that y ' > 0 when
0 < x < 2pi/3 or 4pi/3 < x < 2pi, for x inside [ 0 , 2pi ]

Answer 19

ok i messed up

Answer 20

increasing on (2pi*k, 2pi/3 + 2pi*k) union (4pi/3 +2pi*k, 2pi + 2pi*k )

Answer 21

we can make one nice solution if we do
it is increasing on (-2pi/3 +2pi*k, 2pi/3 + 2pi*k )
and decreasing on
(2pi/3 + 2pi*k , 4pi/3 +2pi*k )