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|dw:1355995034204:dw|

2sin(theta)^2+square root of 3 times sine (theta)=0

\(2\sin^2 x+\sqrt{3sin x}=0\)
right ?>

for some reason tis site is not letting use the equation button

\[2 \sin^2(\theta) + \sqrt{3 \cdot \sin(\theta)} = 0\]

Reload your page.

\[2sine(\theta)^{2}+\sqrt{3}sine(\theta)=0\] i got for theta 0, 180, and 300

sin x(2sin x+sqrt 3)=0
sin x = 0 ---- > x=0,180,360
sin x =-sqrt 3/2 ----> x=.... ?

what? i am, lost

i got -60

\[\sin x(2\sin x+\sqrt{ 3})=0\]
Here just put the factors equal to 0 separately..

0 and 360 is the same exaxt degree. there suppose to be 4 solutions

\[\sin(-60) = - \frac{\sqrt{3}}{2}\]
Yes you are right..

then you add 360. you get back to 300.

yes, u get 2 solutions for sin x = -sqrt 3/2

Generally you can find them using :
\[x = n \pi + (-1)^n y\]

whar are the 2 solution? i only can get 3. how u get the 4th one?"

See here :
\[\sin(x) = \sin(-\frac{\pi}{3})\]

So :
According to that general form :
\[x = n \pi + (-1)^n (-\frac{\pi}{3})\]

i domt see anything o.o

For example :
Put n = 0 here :
\[x = - \frac{\pi}{3}\]

Again refresh your page so see here LATEX..

i did

man i onmly got 3 degrees

What you got as your solutions tell me?

0, 180 and 300

Wait, put n= 2 there and tell me what are you getting??

Is there any range given to you in which you are to find these solutions??
Like 0 to 360..

Are you not getting 240 as one of its solutions?

Not interested in replying ?? @Venomblast

that what im looking for. i got 0 which is the same as 360 degrees and 180

openstudy didnt want to put up my post for like 8 min

Is there any range in which you are to find the solutions or not??

Yes I am telling you the answer..

then what is it?

just tell me. i have to hand this in.

im not use to thqt formula. you learn it differently from me

Sorry, if you want help then I can help you..
So, try to understand what I am saying to you..

i dont understand it

i did do it on my own. dude. i got 3 answes out of 4.i havnt slept yet and my next class is at 8 am

I am just saying you to just put n = 2 here :
\[x = n \pi + (-1)^n (- \frac{\pi}{3})\]

See, if your next class is at 8 am, then it is not my fault..
I can't do anything in this...

i never seen this formula in my life.

But you will see surely and soon..

how do u know. i leanr it differenltly. i cutally have to draw the circle

Now, simply put n = 1 in the formula to get the required solution..

woa.. but i told you already. i already got 300 as my asnwer|dw:1355998903609:dw|

Nice answer..
Well Done..

woa.. but i told you already. i already got 300 as my asnwer

this website need to be better

@waterineyes
Good for nothing.
Ha Ha Ha...

I just showed you, mathematically How you got 300..

ok. but how you get the rest. i just need 1 more degree

Dear, put n = 1 there in the formula..
Why aren't you plugging n = 1 there ??

bnut that will jsut get me 300no?

@waterineyes
You are not helping
Ha Ha Ha,...

No, that will not give you 300..
Try that once.

Yeah..
But I am still trying..

He has done everything, but he does not want to put n = 1 there..

\[x = (2) \pi + (-1)^2 (-\frac{\pi}{3})\]
put n=1
@Venomblast

2pi/3

\[x = (1) \pi + (-1)^1 (-\frac{\pi}{3})\]

-2pi/3

\[x = \pi + \frac{\pi}{3} = ??\]

4/3 pi *

4*180/3=

i need the degree. not the redian.

nvm i will just ask someon eelse.

240 degree.

I am just saying convert that radian into degrees..

that maa work. why cant u justuse the circle.

@Venomblast
what ?

@waterineyes
I am useless lol :D
Not helpful :P

how u got 240. did you add 180 or something? becuase it has to be in quad 3 right. but how?

See :
\[\frac{\pi}{3} = 60\]
Right ??

i had -60. so i added 360. and i got 300 degrees. so now how did you get 240?

It is in 3rd quad.
|dw:1356000009002:dw|

sorry i dont use radians to figure this problem. i told u . i use the actualt cricle

|dw:1356000136227:dw| but how you got 240?

Tell me sine is negative in which quadrants??

@waterineyes
Let me summarize trig
Ha Ha Ha...

|dw:1356008695751:dw|

|dw:1356009445331:dw|

wow, that was almost too painful to read ;)

@amistre64
lol :P