Question

anyone familiar with this: sin(x) - 2sin(x)cos(x) = 0

Answer 1

You need to factor this and then set each factor equal to 0. So set it up like this:
\[\sin(x) - 2\sin(x)\cos(x) = 0 \implies \sin(x)(1-2\cos(x)) = 0\]

Answer 2

Solve the 2 equations:
sinx = 0
1- 2cosx = 0

Answer 3

hey its you!!

Answer 4

still having a hard time resolve this i got 0, pi, 2pi and pi/3

Answer 5

Did your problem give you a restriction? Like did it say \(0 \le\theta \ <\ 2\pi\) or did it not give any restrictions?

Answer 6

(a) Solve for all radian solutions. Give all answers as exact values in radians. Do not use a calculator. (Enter your answer in the form a + bk, where a is in [0, 2π), b is the smallest possible positive number of radians, and k represents any integer. Enter solutions from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes.)

Answer 7

yeah

Answer 8

Alrighty then, so all solutions basically. So since it wants [0,2pi), we can kick out your 2pi answer. So you have 0, pi, pi/3. Youre missing one :)

Answer 9

5pi/3

Answer 10

??

Answer 11

yay!! i got it right thank you

Answer 12

Lol, okay, cool. Good job :)

Answer 13

well now that your here can you help me with some more??

Answer 14

Whatcha got?

Answer 15

2sin^2(theta)-2sin(theta)-1=0 .... i got sin = 1/2, 1

Answer 16

so 30 degrees, and 90??

Answer 17

so since sin is positive its btwn 0 and 180

Answer 18

i put down 30 and 150 and it was wrong...

Answer 19

Yeah, your answers arent going to be quite that pretty, this equation doesnt factor so nicely.
\[2\sin^{2}( \theta ) - 2\sin (\theta) - 1 = 0\] So I'd be plugging these numbers into the quadratic formula to see what we get.
\[\frac{ 2 \pm \sqrt{(-2)^{2} - 4(2)(-1)} }{ 2(2) } = \frac{ 2 \pm \sqrt{12} }{ 4 } = \frac{ 2 \pm 2\sqrt{3} }{ 4 } = \frac{ 1 }{ 2 } \pm \frac{ \sqrt{3} }{ 2 }\]

Answer 20

wow, cant you do (2sin-1) (sin-1)

Answer 21

That would give you:
\[2\sin^{2} ( \theta ) - 3 \sin ( \theta) + 1\]
Did you type the correct equation above? Maybe you mistyped it and thats why I have an odd result?

Answer 22

nope, your right.. ok so I need to use the quadratic equation

Answer 23

Right. Which of course gives you some awkward answers.
\[\sin^{-1} (\frac{ 1 }{ 2 }+\frac{ \sqrt{3} }{ 2 })\]
\[\sin^{-1}(\frac{ 1 }{ 2 }- \frac{ \sqrt{3} }{ 2 })\]

Now the question is are these both valid answers or no?

Answer 24

please tell me i dont have to put each of them into an equation

Answer 25

Well, you wouldnt be able to simplify those answers. I'm just making sure you know why they would or would not be valid answers.

Answer 26

no i dont think they are valid..

Answer 27

oh wait they are

Answer 28

Yes, they are. The domain for sin^(-1)x is [-1,1], so as long as what I plug in is in that interval, they work.

Answer 29

im suppose to come up with a degree so should i just put it in the calculator?

Answer 30

I suppose so. Can't really do anything else.

Answer 31

if it is bigger than 1 then its not possible, right?

Answer 32

If the input is larger than 1 then yes, its not possible.

Answer 33

yay! thank you... i might just keep you here like my security blanket... what are you doing awake right now anyways?

Answer 34

Im often awake at this hour. Unless Im just super tired, its hard for me to be in bed before midnight. Not sure how late ill be up, maybe another hour or two.

Answer 35

wow! i usually am in bed by now... but Im behind in my hw... and I am trying to avoid normal life... so here I am doing hw

Answer 36

I always tryto avoid normal life, haha. I should be doing hw but....I suck at that. I never got much of a sufficient answer for the last question I posted, so Im wonderng if I should just close it and ask a different one.

Answer 37

thats what works for me... keep nagging until you get results

Answer 38

Ill probably just close my current question and ask a new one. Otherwise Im just on here helpng out until im ready to sleep xD

Answer 39

brilliance should be shared :)

Answer 40

I hate to think of myself that way. Im just done a lot of math and have been able to understand a lot of it. I still have to study just like anyone else and work hard. Trig Ive just had enough practice with over time.

Answer 41

just take a compliment!

Answer 42

Fine xD Thanks :P

Answer 43

ok so I need help again.. I dont know if my brain is processing less since I am getting more and more tired. 2cos^2(theta)-2cos(theta)-1=0 I thought it would be 120 and 240

Answer 44

Oh, now we switched to the same question but cosine. So it's going to be the same thing as before. We had to do quadratic formula to get results. And since theyre the SAME coefficients, the quadratic formula will give the same results. So youcan actually take the same answers we got from the previous question, the whole 1/2 +/- sqrt{3}/2 and take inverse cosine of those instead of inverse sine like we did last time.

Answer 45

h=-16t^2+vtsin(theta)
Give the equation for the height, if v is 1,500 feet per second and θ is 30°.
h = -16t2+ t

Answer 46

so how exactly would I insert it... would I just insert -16t^2+1500tsin(pi/3)

Answer 47

UNfortunately got all those stupid glitchy question marks, so not sure what Im missing. But yeah, I'd say plug in just as you did. And Ill assume it wants you to solve for t afterward but yeah, that looks fine :)

Answer 48

ok I am stuck again... ok so here goes one problem... cos(A-85 deg)=sqrt(3)/2

Answer 49

so I knew it would be 30 and 330.... so I added 85 6o 30.. and got 115 + 360. then I added 85 to 330 so that it would be 55 + 360...

Answer 50

Right, so 115 + 360k and 55 + 360k, that works :3

Answer 51

oh i know what i did wrong...

Answer 52

thanks....

Answer 53

Was there something wrong with that answer?

Answer 54

i didnt insert it right... so I am stuck on two more problems then I get to go to sleep

Answer 55

Lol, cool cool.

Answer 56

i am officially on the last problem of this assignment

Answer 57

w00t, lol. Hopefully not a hard one.

Answer 58

im done!!! yay!! i only have 4 more assignments!!! :(

Answer 59

For the rest of thesemester or what?

Answer 60

noooo I have to turn in tomorrow

Answer 61

Eeek, that far behind? o.o

Answer 62

yes!! and Im sick... please tell me you will be here tomorrow... you are the smarted guy here

Answer 63

Ill be here. I cant say when Ill be on, but I will be.