Question

Solve on the interval [0,2Pi):
A. Pi/3 , 2pi/3
B. Pi/6, 5pi/6
C. Pi/3, 5pi/3
D. 2pi/3, 4pi/3

Answer 1

Equation is 1 +sin theta = (sqrt(3) + 2)/2

Answer 2

@CliffSedge

Answer 3

start with
\[\sin(\theta)=\frac{\sqrt{3}}{2}\] and then it should be straight forward

Answer 4

there is a plus 2 on top though?

Answer 5

\[\frac{\sqrt{3}+2}{2}=\frac{\sqrt{3}}{2}+\frac{2}{2}\]

Answer 6

er im getting confused

Answer 7

did you understand my second post?

Answer 8

why did you seperate it?

Answer 9

so show you that the number on the right was identical to
\[\frac{\sqrt{3}}{2}+1\] so that when you subtract 1 from both sides as a first step you get
\[\sin(\theta)=\frac{\sqrt{3}}{2}\]

Answer 10

there isn't a plus one on that side...

Answer 11

\[a+b=c+a\]\[a-a+b=c+a-a\]\[b=c\]It's just one of those rules

Answer 12

wah? what does that have to do with it..

Answer 13

This is your equation:\[1+\sin \theta=\frac{ \sqrt 3 }{ 2 }+1\]Treat one as a in my above response. That's what I meant--since you are adding the same number to both sides, you can subtract that same number from both sides to make the equation into simply sine of theta equals one half the square root of three

Answer 14

there is no plus one on the right .... its just a divide 2...

Answer 15

Did you see where satellite showed you how to separate the fraction giving a +2/2 at the end?

Answer 16

no .i guess my pc being stupid? i dunno

Answer 17

wait.... the +2 is seperate soyou can just cancel it out?

Answer 18

Let's start over.
\[\large Given \space \space 1 +\sin \theta = \frac{\sqrt{3} + 2}{2}\]
\[\large 1 +\sin \theta = \frac{\sqrt{3}}{2} + \frac{2}{2}\]
\[\large \rightarrow 1 +\sin \theta = \frac{\sqrt{3}}{2} + 1 \]
\[\large \rightarrow \sin \theta = \frac{\sqrt{3}}{2}\]

Answer 19

Do you understand all the steps I took to simplify that equation?

Answer 20

yes

Answer 21

now what?

Answer 22

What angle has a sine of √(3)/2?

Answer 23

i can't figure out how to solve for the angle

Answer 24

Consider some common right triangles.

Answer 25

3 /4 /5
?

Answer 26

Ok, what others do you know?

Answer 27

Sine is opposite-over-adjacent.

Answer 28

eh...idon't know any others tbh ...thats the only one i can think of right off the top of my head

Answer 29

oh okay so thats a 30?

Answer 30

i know the angle thing...

Answer 31

On a calculator, you use the inverse-sine function

\[\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60º \]

Answer 32

Now all you need to do is find the other angle that also has that sine on the interval you were given (and convert the angles to radians).

Answer 33

interval ?

Answer 34

what i mean is sin on an interval..its range of numbers?

Answer 35

Trig functions are periodic (they repeat their values regularly). On the interval for Θ given [0,2π), there are two angles that have that same sine ratio.

Answer 36

i can't graph it...
how to graph it...

Answer 37

Do you need to graph it?
You can look at a unit circle to help find the other angles.
Try http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 38

I understand up to sin theta = sqrt3/2

Answer 39

@CliffSedge sorry to bother you with this

Answer 40

What's up?

Answer 41

I need to know how to find the angle and such..I am not sure i even understand the problem...._. can you perhaps do a step by step with me :/ @CliffSedge

Answer 42

@hartnn help me please ._.

Answer 43

'I understand up to sin theta = sqrt3/2' sure ?

Answer 44

Did you look at the problem?

Answer 45

yes, sin theta =sqrt 3 /2 is preety straight forward

Answer 46

now for what theta is sin ratio = sqrt 3/2 ?

Answer 47

yeah i guess i don't understand thequestion really ._.

Answer 48

look for what angles is sin theta = root3/2

Answer 49

30 degrees?

Answer 50

sin 30 = 0.5

Answer 51

and? ...

Answer 52

so its not 30 lol!
:P

Answer 53

but there was a rt3/2
next to it?

Answer 54

SEE THE Y-CO-ORDINATES , NOT X

Answer 55

Omg...DUH! xD

Answer 56

120 ? :3

Answer 57

yes, and in 1st quadrant ?

Answer 58

yeah that would be in the first quadrant

Answer 59

in first quad its 60

Answer 60

wait....
why we going to that?

Answer 61

pi/3 and 2pi/3