Question

I need help! : Solve each of the following. Please give both the general solution and the specific solutions in the given interval. 2 sin (2 theta) + sqrt3 = 0. intervals (0,2pi)

Answer 1

First step?

Answer 2

Hmm...... im not totally sure

Answer 3

We want to isolate theta. So we first move that square root of 3 term over by subtracting it from both sides.


\[\Large 2\sin\left(2\theta\right) + \sqrt{3} = 0\]


\[\Large 2\sin\left(2\theta\right) = -\sqrt{3}\]

Answer 4

What's next?

Answer 5

Ok then would we divide by 2...?

Answer 6

good

Answer 7

\[\Large 2\sin\left(2\theta\right) + \sqrt{3} = 0\]


\[\Large 2\sin\left(2\theta\right) = -\sqrt{3}\]


\[\Large \sin\left(2\theta\right) = -\frac{\sqrt{3}}{2}\]

Answer 8

Next?

Answer 9

we...divide by sin? or 2?

Answer 10

im not really sure

Answer 11

that's a good guess, but that's not it

Answer 12

sine is a function

Answer 13

so we need to apply the inverse function to get rid of it

Answer 14

the inverse of sine is either called "inverse sine" or "arcsine"

Answer 15

Ok

Answer 16

so take the arcsine of both sides...


\[\Large 2\sin\left(2\theta\right) + \sqrt{3} = 0\]


\[\Large 2\sin\left(2\theta\right) = -\sqrt{3}\]


\[\Large \sin\left(2\theta\right) = -\frac{\sqrt{3}}{2}\]


\[\Large 2\theta = \arcsin\left(-\frac{\sqrt{3}}{2}\right)\]

Answer 17

Ok cool, thats what i was thinking

Answer 18

then what would be the nxt step?

Answer 19

Now use this unit circle

http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png

to evaluate the arcsine of \(\Large -\frac{\sqrt{3}}{2}\)

Answer 20

is it 4pi/3?

Answer 21

and 5pi/3

Answer 22

can you explain how to find that?

Answer 23

You found the correct angles, you found the angles x such that

\[\Large \sin(x) = -\frac{\sqrt{3}}{2}\]

Answer 24

Ok what do I do then, like how to formulate that info into the equation?

Answer 25

You can

A) Use the unit circle

B) Memorize the angles

C) Use a calculator

to find the arcsine of a value

Answer 26

I like option A because it allows for exact answers and you don't have to memorize anything.

Answer 27

ok what would be the nexts tep

Answer 28

So you found that the arcsine of \(\Large -\frac{\sqrt{3}}{2}\) is 4pi/3 or 5pi/3

Answer 29

yes

Answer 30

This means we get...


\[\Large 2\sin\left(2\theta\right) + \sqrt{3} = 0\]

\[\Large 2\sin\left(2\theta\right) = -\sqrt{3}\]

\[\Large \sin\left(2\theta\right) = -\frac{\sqrt{3}}{2}\]

\[\Large 2\theta = \arcsin\left(-\frac{\sqrt{3}}{2}\right)\]

\[\Large 2\theta = \frac{4\pi}{3} \ \text{or} \ 2\theta = \frac{5\pi}{3}\]

Answer 31

Ok makes sense

Answer 32

Now because any coterminal angle to 4pi/3 or 5pi/3 works, this means that we can add multiples of 2pi to each angle

Answer 33

ok

Answer 34

\[\Large 2\sin\left(2\theta\right) + \sqrt{3} = 0\]

\[\Large 2\sin\left(2\theta\right) = -\sqrt{3}\]

\[\Large \sin\left(2\theta\right) = -\frac{\sqrt{3}}{2}\]

\[\Large 2\theta = \arcsin\left(-\frac{\sqrt{3}}{2}\right)\]

\[\Large 2\theta = \frac{4\pi}{3} \ \text{or} \ 2\theta = \frac{5\pi}{3}\]

\[\Large 2\theta = \frac{4\pi}{3}+2\pi k \ \text{or} \ 2\theta = \frac{5\pi}{3}+2\pi k\]

Answer 35

where k is any integer

Answer 36

Last step?

Answer 37

divid 2?

Answer 38

so theta is isolated?

Answer 39

yes, divide everything by 2

\[\Large 2\sin\left(2\theta\right) + \sqrt{3} = 0\]

\[\Large 2\sin\left(2\theta\right) = -\sqrt{3}\]

\[\Large \sin\left(2\theta\right) = -\frac{\sqrt{3}}{2}\]

\[\Large 2\theta = \arcsin\left(-\frac{\sqrt{3}}{2}\right)\]

\[\Large 2\theta = \frac{4\pi}{3} \ \text{or} \ 2\theta = \frac{5\pi}{3}\]

\[\Large 2\theta = \frac{4\pi}{3}+2\pi k \ \text{or} \ 2\theta = \frac{5\pi}{3}+2\pi k\]

\[\Large \theta = \frac{4\pi}{6}+\pi k \ \text{or} \ \theta = \frac{5\pi}{6}+\pi k\]

\[\Large \theta = \frac{2\pi}{3}+\pi k \ \text{or} \ \theta = \frac{5\pi}{6}+\pi k\]

Answer 40

ok is that my final?

Answer 41

yes, those are your general solutions

Answer 42

wait if you divide by 2 wouldnt the 5pi/6 change?

Answer 43

and the 2pi/3??

Answer 44

I did that. I started with 4pi/3 and divided by 2 to get 4pi/6. Then reduced to get 2pi/3

Answer 45

Then I divided 5pi/3 by 2 to get 5pi/6

Answer 46

Oh ok:) thx, you are really the best. do u mind if i ask 2 more? i can tell you what i wpuld do next so youre not doing everything? sound good
?

Answer 47

wait a sec

Answer 48

those were the general solutions

Answer 49

yeah

Answer 50

they now want the "specific solutions in the given interval"

Answer 51

and the given interval is (0,2pi)

Answer 52

Ok, im not 100% sure how to find that

Answer 53

So how do we find these solutions?

Answer 54

the specific

Answer 55

hmmm.......

Answer 56

Here's the best way to do find the solutions in the given interval

Start with

\[\Large \theta = \frac{2\pi}{3}+\pi k \]

and plug in k = 0

What do you get?

Answer 57

I get theta=2pi/3 + 0

Answer 58

good, is that in the interval (0,2pi) ???

Answer 59

yes?

Answer 60

you can check

2pi = 6.283185
2pi/3 = 2.094395

So 2.094395 is definitely between 0 and 6.283185

Answer 61

Therefore, 2pi/3 is between 0 and 2pi

Answer 62

ok!

Answer 63

now plug in k = 1 and tell me what you get

Answer 64

I get theta=2pi/3 + pi

Answer 65

Ok, add those terms

Answer 66

I got theta= 5pi/3

Answer 67

good, now check to see if its in range

Answer 68

it is i believe

Answer 69

sry i dont have a calc near me right now

Answer 70

5pi/3 = 5.23598775598299


which is less than 6.283185, so yes it is

Answer 71

Now plug in k = 2 and simplify

Answer 72

ok

Answer 73

i get theta=8pi/3

Answer 74

?

Answer 75

8pi/3 = 8.37758040957279, is that in range?

Answer 76

no

Answer 77

o r is it?

Answer 78

you are correct, it's not in range

Answer 79

ok

Answer 80

so then what would my specific solution answers be?

Answer 81

So far, we found that k = 0 and k = 1 work (but k = 2 does not), correct?

Answer 82

right

Answer 83

these k values produce theta values of 2pi/3 and 5pi/3

Answer 84

Yes

Answer 85

so those two solutions are specific solutions (so far..)

Now we must check \[\Large \theta = \frac{5\pi}{3}+\pi k \]

Answer 86

ok

Answer 87

If you plug in k = 0, you get 5pi/3

if you plug in k = 1, you get 8pi/3 (I'm sure you see the pattern)

So the solutions overlap...which means that there are only 2 solutions in this interval

Answer 88

ok

Answer 89

and those two solutions are 2pi/3 and 5pi/3

Answer 90

ok

Answer 91

so 2pi/3 and5pi/3 are sp[ecific soliutions

Answer 92

specific solutions

Answer 93

you got it

Answer 94

Ok! do u not mind helping with just two more?

Answer 95

they are specific solutions to \(\Large 2\sin\left(2\theta\right) + \sqrt{3} = 0\) in the interval \(\Large \left(0,2\pi\right)\)

Answer 96

sure, go for it

Answer 97

Ok here is the next one: same directions as the first. the problem is:

2 sin^2 x = 2 + cos x --> interval (0,pi)

Answer 98

ok, do you know any trig identities?