Solve the interval [0,2pi): 1+sin theta=1/2 A.2pi/3, 4pi/3 B.pi/6, 5pi/6 C.pi/3, 5pi/3 D.7pi/6, 11pi/6

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Solve the interval [0,2pi): 1+sin theta=1/2 A.2pi/3, 4pi/3 B.pi/6, 5pi/6 C.pi/3, 5pi/3 D.7pi/6, 11pi/6

Mathematics
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subtract 1 from both sides of the equation: sin theta = -1/2
Do I graph sin theta =-1/2?
What does the [0,2pi) even mean?

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[0,2pi) means the solutions in the range 0 to 2pi radians
So can that limit my answers, or it doesn't matter?
first find the reference angle use your calculator to find the angle whose sine is 1/2 enter sin-1 (1/2) make sure that your calc is set to radians can you do that?
yes it limits the answers to angles in that range
Yeah, althoght dont I need the unit circle for that, and can graphing be another way to solve the problem?
you can but its easier to use calc or you can use the dtandard triangle below:- |dw:1437850039490:dw|
When I type in arcsin (1/2) I get pie/6? Is that correct?
yes - and you see its pi.6 from the above trianggle
Yeah, but does that mean that its a solution to the problem?
but the sine is -1/2 and the angles with this sine are in the 3rd and 4th quadrants
|dw:1437850196121:dw|
oh - ive written them in degrees 30 degrees = pi/6 radians
Oh!!!!, so now what?
|dw:1437850294435:dw|
thats one angle: 7pi/6 can you work out the other?
I think so, let me trie! :0 By the way thank for yoir help, really appreciated!
one revolution = 2pi radians
yw
I don't know if its correct but I get 11pi/6?
thats correct 2pi - pi/6 = 12pi/6 - pi/6 = 11pi/6
D is correct choice
Okay thanks again! :)
yw

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