Question

what values for Ø(0 ≤Ø ≤ 2 pi ) satisfy the equation 3 sin Ø= sin Ø-1

Answer 1

@goformit100 @geekfromthefutur @ofmice&men

Answer 2

@foreverjennnicole @no;biuvjh,mgj

Answer 3

3sina=sina-1
2sina=-1
sina=-1/2
sinüs is minus in third and fourth quadrant and sin60=1/2
so 60+180=240 and 360-60=300 degrees are the answers.

Answer 4

the possible answers are

a: pi/3 , 5pi/3

b: 2pi/3 , 4pi/3

c: pi/6 , 3pi/6

d: 7pi/6 , 11pi/6

Answer 5

I made a terrible mistake sin60=3^(1/2)/2 sin30=1/2 :(
Thus 180+30=210 and 360-30=330 are the answers so it is d :)

Answer 6

thank you QAQ would you mind sticking around and helping me with the rest?

Answer 7

ı can see them

Answer 8

thank you so much!

Answer 9

the next one is the same but with a different equation. tan^2 Ø = -3/2 sec Ø and the possible answers are

a: pi/2 , 5pi/2

b: 3pi/4 , 5pi/4

c: pi/3 , 5pi/3

d: 2pi/3 , 4pi/3

Answer 10

@jim_thompson5910

Answer 11

on this page

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

you'll see that

\[\Large \tan^2(\theta) + 1 = \sec^2(\theta)\]

(see page 2)

Answer 12

solve for tan^2 to get

\[\Large \tan^2(\theta) = \sec^2(\theta) - 1\]

Answer 13

Then plug that into the equation to get

\[\Large \tan^2(\theta) = -\frac{3}{2}\sec(\theta)\]

\[\Large \sec^2(\theta) - 1 = -\frac{3}{2}\sec(\theta)\]

Answer 14

do you know how to solve from here?

Answer 15

no Q_Q i dont know how to do any of this thats why i'm here

Answer 16

ok let z = sec(theta)

that means z^2 = sec^2(theta)

Answer 17

and that allows us to replace every copy of sec(theta) with z to get

z^2 - 1 = (-3/2)*z

Answer 18

are you able to solve z^2 - 1 = (-3/2)*z for z?

Answer 19

no... sorry...

Answer 20

z^2 - 1 = (-3/2)*z

2z^2 - 2 = -3z ... multiply everything by 2 to clear out the fraction

2z^2 - 2 + 3z = 0

2z^2 + 3z - 2 = 0

Now use the quadratic formula to solve for z

Answer 21

how would this get any of the possible answers? the possible answers are all fractions with pi

Answer 22

we'll get there, don't worry

Answer 23

what solutions do you get in terms of z?

Answer 24

i dont know the quadratic formula...

Answer 25

the formula (and examples using it) is given here

http://www.purplemath.com/modules/quadform.htm

Answer 26

in this case, a = 2, b = 3, c = -2

Answer 27

1.5/4 and 7.5/4

Answer 28

You should get

\[\Large z=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[\Large z=\frac{-3 \pm \sqrt{(3)^2-4(2)(-2)}}{2(2)}\]
\[\Large z=\frac{-3 \pm \sqrt{25}}{4}\]
\[\Large z=\frac{-3 \pm 5}{4}\]
\[\Large z=\frac{-3 + 5}{4} \ \text{ or } \ z=\frac{-3 - 5}{4}\]
\[\Large z=\frac{2}{4} \ \text{ or } \ z=\frac{-8}{4}\]
\[\Large z=\frac{1}{2} \ \text{ or } \ z=-2\]

Answer 29

if z = 1/2, and z = sec(theta), then

z = 1/2

sec(theta) = 1/2

theta = ????

Answer 30

i dont know Q_Q

Answer 31

use the identity sheet I gave you

secant is what in terms of cosine?

Answer 32

i looked but i cant find it...

Answer 33

look under reciprocal identities

Answer 34

near the top of page 2

Answer 35

sec=1/cos???

Answer 36

correct

Answer 37

so if sec(theta) = 1/2, then taking the reciprocal of both sides gives

1/sec(theta) = 2/1

cos(theta) = 2

is it possible for cos(theta) = 2 ?

Answer 38

yes?

Answer 39

actually no it's not possible because the range of cosine is -1 to 1

1 is the largest value cosine can be

Answer 40

cos(theta) = 2 has no solutions

Answer 41

if z = -2, then...

z = -2

sec(theta) = -2

1/cos(theta) = -2

cos(theta) = -1/2

use a unit circle to find the value of theta that makes that equation true

Answer 42

i still dont understand any of this

Answer 43

so none of this makes any sense?

Answer 44

i'm trying to understand but...

Answer 45

if so, then perhaps you're in the wrong lesson?

Answer 46

this class is mandatory... its required to graduate.. and they went and stuffed difficult trig into an algebra class

Answer 47

the class is already so difficult that i am FAILING. i understand none of it and my teacher teaches all of the other math classes for the school so shes literally useless

Answer 48

completely unfair since trig and algebra are somewhat different classes

Answer 49

there's no way to transfer?

Answer 50

nope. not that my mom would let me anyways. she doesnt care. all she does is yell at me and tell me to "do better"

Answer 51

so since this is an algebra class (I thought it was a trig class), I'm guessing this is the first time you've seen the quadratic formula?

Answer 52

they had the quadratic formula in algebra 1 but i havent used it since so i had forgotten what it was

Answer 53

ok, that's good it's not completely new

as for the trig stuff, all of it seems completely foreign?

Answer 54

hopefully you've seen some of the identities before

Answer 55

all of the trig is new... and sadly i have another quiz and a test after the quiz i am working on... and i have to get 80 or above on all of them...

Answer 56

@jim_thompson5910 you still there?

Answer 57

yeah sorry I got distracted and I just noticed the notification

as a rule we can't help with quizzes, tests, or exams of any kind (since they must be done on your own). But you can go to the hw, lectures or study guides to work on similar problems. We can help with those. Working on them will help you practice and get ready for the tests.

Answer 58

When did they introduce all of this new trig? At the last minute so to speak?

Answer 59

we are in unit 6 and this short unit is entirely trig so yes it was very last minute

Answer 60

unit 6 is the last unit? I guess that would make sense (in a way) because they're trying to transition you from algebra to trig

Answer 61

the transition is a bit too fast though

Answer 62

how much do you have in terms of hw? or lecture material?

Answer 63

it isnt the last unit. we have a couple units after this and, according to the names of the lessons, we are going back to things that i learned in algebra 1. so this unit is completely random

Answer 64

for this particular unit

Answer 65

hmm weird

Answer 66

it's like they just gave up trying and clobbered together a whole bunch of things

Answer 67

it is making it impossible for me to get my grade up enough so that i can pass and graduate (i am in my senior year of high school and i dont understand any of this)

Answer 68

which is why I was hoping transferring was possible (so you can get into a class that actually teaches trig if it's a trig class...not a weird mix)

Answer 69

anyways, I might as well finish up since I got this far

so we have cos(theta) = -1/2

we use the unit circle

http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/720px-Unit_circle_angles_color.svg.png

to find values of theta that make cos(theta) = -1/2 true. Basically we are looking for points on the unit circle that have x coordinates of -1/2

According to the unit circle, cos(120) = -1/2 and cos(240) = -1/2

So theta = 120 and theta = 240

those angles are in degree mode

The unit circle provided also shows radians as well. Because 120 degrees = 2pi/3 radians and 240 degrees = 4pi/3 radians, this means the solutions to the original equation are 2pi/3 & 4pi/3

There are infinitely many solutions, but remember theta is restricted to be in the interval from 0 to 2pi