Question

@callisto, can you check my answer?

Answer 1

Ok so do you remember question a?

Answer 2

Can you post it again?

Answer 3

This is what we did to solve it:
first, I have to factor out the quadratic equation by decomposition method
6x^2 + 3x - 2x -1 = 0
(6x^2 + 3x) (-2x -1)
3x(2x + 1) -1(2x+1)=0
(3x-1)(2x+1)=0
Then,we write 2 equations
(2cosx + 1) = 0-----(1)
OR
(3cosx - 1) =0-----(2)

We isolate the trig function, Which gives you cosx = -1/2 and cosx = 1/3.

for equation 1, the answer is 120degrees and 240degrees (we can look at unit circle)

For equation 2, we were not sure how you got 70.52878 or 289.4712

Answer 4

Is my procedure correct for equation 1? and can you show me how to solve equation 2?

Answer 5

Yup, question 1 should be correct

Answer 6

equation 1 is correct? How do we solve the equation 2 to get what you got?

Answer 7

For question (2), you need to change the sin into cos first. Can you try it?

Answer 8

Wait, before we move on to the next one, Can you show me how yo get equation 2?

Answer 9

Its just that i need to be able to explain every step of the procedure

Answer 10

you mean (b) right?

Answer 11

no, If you see in my answer

Answer 12

There are 2 equations, equation 1 and equation 2

Answer 13

equation 1 = 120degrees and 240degrees
equation 2 = 70.52878 or 289.4712

I know how to get equation 1, But i have trouble remembering how to solve equation 2

Answer 14

So i wanted to know how you solved (3cosx - 1) and got 70.52878 or 289.4712

Answer 15

Okay
(3cosx - 1) =0
Both sides add 1 and you'll get
(3cosx - 1) +1=0 +1
3cosx =1
Got this part?

Answer 16

Yes

Answer 17

Divide both sides by 3, what would you get?

Answer 18

I would get cos(x) = 1/3

Answer 19

or if its decimal, .3333

Answer 20

As the normal practice, you would use the arc cos to find the angle, what would you get?

Answer 21

i would get 70degrees

Answer 22

How about the 289?

Answer 23

There are 2 quadrants which gives positive value for cos, what are they?

Answer 24

they are q4 and q1

Answer 25

Yup, for the angle in quadrant I =x, the following would give you the same absolute value to a trigo ratio
angle in quadrant II = 180 -x
angle in quadrant III = 180 +x
angle in quadrant IV =360 -x

So, can you work out the angle in quadrant IV?

Answer 26

360 - 70. Okkk

Answer 27

that makes sense:).
Thank you for explaining that to me:)
But my procedure and answer are correct right?

Answer 28

Yes

Answer 29

Just a minute, sorry!!

Answer 30

perect!:)

Answer 31

Sure, I will wait.

Answer 32

Perhaps we can start (b) now?!

Answer 33

Welcome Back:) Yes

Answer 34

When you said, change sin into cos, do you mean 1st pythagorean identity?

Answer 35

1- sin^2 x =?

Answer 36

cos² x

Answer 37

Cool~ can you rewrite the equation now?

Answer 38

so is it cos² x = cos x?

Answer 39

nope!!!! Expressing the equation in only one kind of ratio can help you solve the equation easier. Otherwise, it's hard to solve

Answer 40

so 2cos² x = 0. is this what you mean?

Answer 41

Nope o.o

Answer 42

Can we start again? suddenly, I've found a faster method :S

Answer 43

That would be great:)

Answer 44

So, consider the LHS only, take out the common factor

Answer 45

What would you get now?

Answer 46

Is it sin²(x)

Answer 47

nope...
\[2-2 sin^2 x =cosx\]\[2(1-sin^2x)=cosx\]
Understand this part?
BTW, what's the time there?

Answer 48

its 3am

Answer 49

My brain is not fully cunctional right now, But I have to finish my hw

Answer 50

Hmm.. let's finish it earlier so that you can sleep earlier!!!
Do you understand the step I wrote?

Answer 51

Good Idea:). Yes

Answer 52

so, 1-sin^2 x would give you cos^2 x, agree?

Answer 53

Agreed

Answer 54

so, replace 1-sin^2 x by cos^2 x in the equation. This gives
\[2(cos^2x) = cosx\], got it?

Answer 55

Yes

Answer 56

so, both sides minus cosx, and you'll get
\[2cos^2x-cosx=0\] Understand?

Answer 57

Yes.

Answer 58

Take out the common factor on the LHS, what would you get?

Answer 59

Is it 2cosx cosx?

Answer 60

Nope....
\[2cos^2x -cosx=0\]\[cosx(2cosx-1)=0\]
Understand? (it's actually doing factorization)

Answer 61

Oh ok

Answer 62

Oh, I thought you left:)

Answer 63

Sorry, was switching the PC with my sister :( actually... she forced me..)

Answer 64

oh. lol

Answer 65

So,
cosx=0 or (2cosx−1)=0
understand?

Answer 66

Yes, understand

Answer 67

Can you solve it from here?

Answer 68

k. one sec

Answer 69

so is cos(x) = 0 = 1/2
and (2cosx-1) = 0 = 1/3?

Answer 70

Nope!!!!!!!

Answer 71

one second

Answer 72

cos(x) = 0 ----(1) (2cosx-1) = 0 ----(2)
First solve (1)

Answer 73

ok

Answer 74

ok, so is cos x = pi/2? if so, then is equation 1 = 90degrees?

Answer 75

90 degree is one of the values, can you work out the other one?
Hint: use 360-x to figure it out :)

Answer 76

so 360-90? 270?

Answer 77

yes!~~~~

Answer 78

haha. I am EXTRA slow when I am tired arent I?

Answer 79

My brain is not even working with me. It is just blank. hahaha. sorry I took forever

Answer 80

No sorry~~~

Answer 81

Thanks you so much for your help:)

Answer 82

I was having a serious headache too~ but it's fine now~ So, do hw, go to sleep and it'll be fine :)

Answer 83

Now,solve (2cosx-1) = 0

Answer 84

ok

Answer 85

Wait, isnt the the same as equation 1?

Answer 86

lol, yes :P

Answer 87

If i get pi/2, then that means I get teh exact same as equation 1

Answer 88

in question (a), i think

Answer 89

so for equation 2, same thing 90degrees and 270degrees

Answer 90

hahaha. While I had to solve and all, you already knew from the beginning didnt you? :)

Answer 91

:P

Answer 92

What do you get for second question?

Answer 93

which second question?

Answer 94

isnt the answer for second question 90degrees and 270degrees for both equations (1, 2)

Answer 95

2- 2sin^2 x = cosx
Can you answer the question in full workings as well as the answer?

Answer 96

Not really, for equation (1), yes, (2), no