Question

find all solutions to the equation in the interval [0,2pi): 2sin^3x + sin^2x = 0

Answer 1

is it\[2\sin^3x+\sin^2x=0\]

Answer 2

yes

Answer 3

did you factor?

Answer 4

i didn't, no

Answer 5

so do that and set each factor equal to zero.

Answer 6

i'm not sure i tried but it doesn't look right

Answer 7

sorry, my computer lost power

Answer 8

no problem :)

Answer 9

\[2\sin^3x+\sin^2=0 \Rightarrow \sin^2x \left( 2\sin x+1 \right)=0\]
so \[\sin^2x = 0\]and\[2 \sin x+1=0 \Rightarrow \sin x = -\frac{ 1 }{ 2 }\]

Answer 10

so what values of x in that interval satisfy those equations?

Answer 11

as in what are the possible answers?

Answer 12

yep

Answer 13

a. 5pi/6, 11pi/6
b. 4pi/3, 5pi/3
c. 0, 7pi/6, 11pi/6
d. 0, pi/2, pi, 4pi/3, 3pi/2, 5pi/3
e. none of these

Answer 14

not like that... you tell me what values of x make \[\sin^2x=0\]and\[\sin x = -\frac{ 1 }{ 2 }\]

Answer 15

well sin^2x=0 would be at 0, 90, 180, 270, and 360 in terms of degrees and sin x = -1/2 would be at 0 and 30

Answer 16

no... be care and check again

Answer 17

sin2(x) = 1/2 - 1/2 cos(2x)?

Answer 18

no... do you know the unit circle?

Answer 19

yes

Answer 20

so when is sin x = -1/2

Answer 21

120 (2pi/3), 210 (7pi/6), 270 (3pi/2), 330 ( 11pi/6)

Answer 22

and sin^2x= 0 at 360 (2pi) right?

Answer 23

not all of thise for sin x = -1/2... you said you knew the unit circle

Answer 24

sin x = -1/2 at 210 and 330 degrees and sin^2x= 0 at 360

Answer 25

you're missing one for sin^2 x and 360 = 2pi which isn't in the interval!

Answer 26

ah. sin^2x=0 180?

Answer 27

yep

Answer 28

ahh, ok

Answer 29

but then from the answer selection: a. 5pi/6, 11pi/6
b. 4pi/3, 5pi/3
c. 0, 7pi/6, 11pi/6
d. 0, pi/2, pi, 4pi/3, 3pi/2, 5pi/3
e. none of these

it would be none of these because sin x = -1/2 is at 7pi/6 and 11pi/6 but sin^2x=0 can't be 360 and would be pi, which ins't a 0 (referring to answer c) and in answer d it's missing 7pi/6 and 11pi/6

Answer 30

yeah, it would have to be 0, pi, 7pi/6 and 11pi/6

Answer 31

alright, thanks a ton man for the help. sorry it took so long for me to understand your explanations at the start

Answer 32

no worries! good luck!

Answer 33

:) you too