Question

http://prntscr.com/9888fc
What am I doing wrong?

Answer 1

If you "check your work",
you'll notice that you're not getting -13 in the middle as you need.

\[\large\rm 3x^2\color{orangered}{-13x}+4\]\[\large\rm 3x^2\color{orangered}{-12x-x}+4\]Factoring a 3x from each of the first two terms,\[\large\rm 3x(x-4)-x+4\]Factoring a -1 from the other two terms,\[\large\rm 3x(x-4)-(x-4)\]Further factoring,\[\large\rm (x-4)(3x-1)\]Any confusion on those steps? :o

Answer 2

what

Answer 3

lol

Answer 4

oh

Answer 5

how did I mess up that big ugh

Answer 6

Probably too busy being a silly puppy wuppy kittiwitti is my guess c:
buhaha!

Answer 7

-.-

Answer 8

Another question: http://prntscr.com/988j5s
@zepdrix ?

Answer 9

zeppy you and your silly jokes crack me up sometimes :U

Answer 10

\[\large\rm \color{orangered}{\sin x}-2\color{orangered}{\sin x} \cos x=0\]More factoring, ya? :)

Answer 11

\[\large\rm \color{orangered}{\sin x}(1-2\cos x)=0\]Understand what to do from that point?

Answer 12

OH
...no.

Answer 13

:c

Answer 14

Apply your Zero-Factor Property:\[\large\rm \sin x=0\qquad\qquad\qquad (1-2\cos x)=0\]Do you know to find the solutions that sin x is producing? :)
What angle gives you a sine of 0?

Answer 15

I mean, the sine of what angle, gives you zero*

Answer 16

0pi, 2pi, pi

Answer 17

o vo

Answer 18

Sounds good.
So our first set of solutions is:\[\large\rm x=0+k \pi,\qquad\qquad k\in\mathbb Z\]The sine of any integer multiple of pi gives us 0.

Answer 19

How bout the other solutions? :)\[\large\rm 1-2\cos x=0\]Can you figure out the angles for this one?

Answer 20

*lost*
s'2 am here ... ;;

Answer 21

I'm in trig and I don't recognize that symbol

Answer 22

the E-thing or the Z

Answer 23

Oh sorry :)

Answer 24

the e thing means in and Z is the set of all integers.
so k is in the set of all integers
hmm actually I don't remember seeing that when I did trig either. >_>

Answer 25

\[\large\rm k=0,1,2,...\]

Answer 26

k is integer values,\[\large\rm \sin(-2\pi)=0\]\[\large\rm \sin(0)=0\]\[\large\rm \sin(\pi)=0\]\[\large\rm \sin(2\pi)=0\]...\[\large\rm \sin(k \pi)=0\]

Answer 27

For the other one,
you need to start by isolating the cos x,
:)

Answer 28

\[\large\rm 1-2\cos x=0\]So ummm,
subtract 1,\[\large\rm -2\cos x=-1\]divide by -2,\[\large\rm \cos x=\frac{1}{2}\]ya? :)
Again, you need to think about which angles this corresponds to.
You should get two of them.

Answer 29

*brain died*

Answer 30

gimme a sec xD

Answer 31

AH
I think i get it

Answer 32

so arcos(1/2)?

Answer 33

I suppose -_-
That's one of the special ones from your unit circle that you need to remember though :o

Answer 34

eh... pi/3?

Answer 35

pi/3 Hmm yes, that sounds right :)
Cosine corresponds to your x-coordinate.
And your x-coordinate is the shorter length, the 1/2, at pi/3.

There is another angle that produces 1/2 though.

Answer 36

This one in the 4th quadrant.

Answer 37

oh

Answer 38

2pi-pi/3=5pi/3

Answer 39

Good good good.

And as we noticed with the sinx,
we can also consider any angle which is `co-terminal` with our angles to also be solutions.

So if we start at pi/3 and spin a full time around the circle,
we end up at the same point, which still gives us 1/2 when we take its cosine, ya?

So pi/3 + 2pi is also a solution
then pi/3 + 4pi is also a solution..