Question

help

Answer 1

@Vocaloid

Answer 2

I don't remember how to do #1 by hand, I'd just put both the equations into a graphing calc/etc. and see how many intersections there are between 0 <x < 2pi

Answer 3

2?

Answer 4

yeah that's what I got too

Answer 5

for # the error is in these two steps, do you see what it is?

Answer 6

b

Answer 7

good

Answer 8

hm. I'm a little rusty on this

tan^2 = sec^2 - 1

so 2(sec^2 - 1) - sec - 1 = 0

I'm not sure where to go from there though :S

Answer 9

@Shadow do you remember how to do this ;;

Answer 10

No @Zarkon ?

Answer 11

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

Answer 12

?

Answer 13

you get that from \[\sin^2(x)+\cos^2(x)=1\]

Answer 14

divide both sides by \[\cos(x)\]

Answer 15

* \[cos^2(x)\] that is

Answer 16

are you just trying to solve \[2(\sec^2(\theta) - 1) -\ sec(\theta) - 1 = 0\]

Answer 17

a and d?

Answer 18

lol...sorry ..didn't see your post of the image...jas

Answer 19

its okay lol am i correct?

Answer 20

?

Answer 21

sec(x)=-1 and sec(x)=3/2

Answer 22

looking at the tangent ones now

Answer 23

only 2 answer choices allowed

Answer 24

then the ones i said

Answer 25

okay

Answer 26

#4?

Answer 27

do you have a guess?

Answer 28

c maybe

Answer 29

\[\frac{\sqrt{3}}{2}\] is not the answer

Answer 30

b?

Answer 31

yes

Answer 32

ty

Answer 33

np