Question

What values for (posted below) satisfy the equation?

Answer 1

the first attachment fills in the blank, the second one follows the sentence

Answer 2

factor out the common factor of cosine first and start with
\[\cos(x)(1-\tan(x))=0\]
then set each factor equal to zero and solve

Answer 3

you good from there?

Answer 4

I have no clue how to factor these problems

Answer 5

i factored it for you

Answer 6

How do I get to the final answer?

Answer 7

each term had a common factor of cosine
don't get hung up on the fact that they are functions
if you wanted to factor say
\[x-xy\] you would no doubt come up with
\[x(1-y)\] right away

Answer 8

suppose the question said "solve for \(x\) and \(y\) \(x-xy=0\)" what would you do?

Answer 9

subtract y

Answer 10

hmm no
\[x-xy=0\\
x(1-y)=0\\
x=0\text {or } 1-y=0\\
x=0,y=1\]

Answer 11

ok im so lost lol sorry

Answer 12

you have the same kind of question here, but instead of \(x\) and \(y\) you have \(\cos(x)\) and \(\tan(x)\)

Answer 13

maybe because you are hung up on the trig? lets forget trig for a moment

Answer 14

i mean we still have to worry about that at the end, but not at the beginning
at the beginning we only need to use the "zero property" or whatever that is called, i.e. factor the expression and set each factor equal to zero
that is step one

Answer 15

yeah thats probably the problem, im learning though lol

Answer 16

so we start by factoring the expression
\[\cos(x)-\cos(x)\tan(x)\] is it clear that each term has a cosine in it?

Answer 17

yes

Answer 18

ok as the math teachers say "factor it out" i.e. write that expression in factored form
what do you get ? (hint: it is the very first thing i wrote above)

Answer 19

\[\cos (x)-\cos (x)\tan (x)\]

Answer 20

that is the expression BEFORE it is factored
factor out the common factor of cosine
then what does it look like ?

Answer 21

\[x-xy=0\]

Answer 22

\[x(1-y)=0\]

Answer 23

\[x=0or 1-y=0\]

Answer 24

ok that was my example with x and y
how about with cosine and tangent?

Answer 25

x=0,y=1

Answer 26

ok now we have it with x and y
repeat with cosine instead of x and tangent instead of y

Answer 27

both are zero?

Answer 28

start with
\[\cos(x)(1-\tan(x))=0\] that is the factored form
then set each factor equal to zero and solve for \(x\)

Answer 29

x=-1?

Answer 30

lets go slow

Answer 31

do you see how i arrived at
\[\cos(x)(1-\tan(x))=0\]?

Answer 32

no, i don't know how to do any of this sadly

Answer 33

opps didnt meant to put that

Answer 34

i am sure you didn't
the first step is algebra, not trig

Answer 35

to find X you have to eliminate all the other factors in the equation so that X stands alone

Answer 36

the first step is to factor
then next step is to set each factor equal to zero and solve
so if
\[\cos(x)(1-tan(x))=0\] then either
\[\cos(x)=0\] or
\[1-\tan(x)=0\]

Answer 37

do you know the values of \(x\) for which
\[cos(x)=0\]?

Answer 38

pi?

Answer 39

not to be mean, but did you just guess?

Answer 40

no, it is not \(\pi\)
\[\cos(\pi)=-1\] not zero

Answer 41

I was looking at the first part of the equation that i put in above, where it says 2pi, but yes it was kind of a guess

Answer 42

lets not guess, lets find it
look at the unit circle on the last page of the attached cheat sheet
you will see lots of coordinates corresponding the the different angles
the first ordinate is cosine, the second is sine

Answer 43

find the two places on the unit circle where the first coordinate is zero
then find angle corresponding to those points
let me know when you see them

Answer 44

\[\frac{ \pi }{ 2}\]

Answer 45

yes!!

Answer 46

that is one of them, there is one more

Answer 47

\[\frac{ 3\pi }{ 2}\]

Answer 48

bingo

Answer 49

yay lol

Answer 50

that means
\[\cos(\frac{\pi}{2})=0\]and
\[\cos(\frac{3\pi}{2})=0\]so those are two of your four answers

Answer 51

now on to tangent
we need to solve
\[1-\tan(x)=0\]i.e.
\[\tan(x)=1\]

Answer 52

please don't post possible answers
we will find them

Answer 53

we already have two of the four
\[\frac{\pi}{2},\frac{3\pi}{2}\]

Answer 54

OH im just posting in general cuz my school is crazy and sometimes the REAL answer is an option

Answer 55

next we need to solve
\[\tan(x)=1\] any ideas? (no is a fine answer)

Answer 56

\[\frac{ \pi }{ 4 } and \frac{ 5\pi }{ 4 }\]

Answer 57

yes, not sure how you got that, but it is right

Answer 58

my brother helped me find that awhile ago, i gave up on this question weeks ago ha,

Answer 59

final answer to your question is four numbers
\[\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{4},\frac{5\pi}{4}\]

Answer 60

\[\frac{\pi}{4}\]and \[\frac{5\pi}{4}\] are the two places one the unit circle where cosine and sine are equal
you can look at your cheat sheet and see it

Answer 61

since
\[\tan(x)=\frac{\sin(x)}{\cos(x)}\] tangent will be one if sine and cosine are equal

Answer 62

yay, thank you so much for helping me figure out the final answer!

Answer 63

yw

Answer 64

C is the correct answer?

Answer 65

\[\frac{ \pi }{ 4} and \frac{ 5\pi }{ 4}\]