What values for (posted below) satisfy the equation?

- anonymous

What values for (posted below) satisfy the equation?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

you good from there?

- anonymous

I have no clue how to factor these problems

- anonymous

i factored it for you

- anonymous

How do I get to the final answer?

- anonymous

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

- anonymous

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

- anonymous

subtract y

- anonymous

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

- anonymous

ok im so lost lol sorry

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yeah thats probably the problem, im learning though lol

- anonymous

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

- anonymous

yes

- anonymous

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)

- anonymous

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

- anonymous

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

- anonymous

\[x-xy=0\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

x=0,y=1

- anonymous

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

- anonymous

both are zero?

- anonymous

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

- anonymous

x=-1?

- anonymous

lets go slow

- anonymous

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

- anonymous

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

- anonymous

##### 1 Attachment

- anonymous

opps didnt meant to put that

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

pi?

- anonymous

not to be mean, but did you just guess?

- anonymous

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

- anonymous

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

- anonymous

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

##### 1 Attachment

- anonymous

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

- anonymous

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

- anonymous

yes!!

- anonymous

that is one of them, there is one more

- anonymous

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

- anonymous

bingo

- anonymous

yay lol

- anonymous

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

- anonymous

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

##### 1 Attachment

- anonymous

please don't post possible answers
we will find them

- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yw

- anonymous

C is the correct answer?

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.