anonymous
  • anonymous
What values for (posted below) satisfy the equation?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
the first attachment fills in the blank, the second one follows the sentence

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anonymous
  • 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
  • anonymous
you good from there?
anonymous
  • anonymous
I have no clue how to factor these problems
anonymous
  • anonymous
i factored it for you
anonymous
  • anonymous
How do I get to the final answer?
anonymous
  • 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
  • anonymous
suppose the question said "solve for \(x\) and \(y\) \(x-xy=0\)" what would you do?
anonymous
  • anonymous
subtract y
anonymous
  • anonymous
hmm no \[x-xy=0\\ x(1-y)=0\\ x=0\text {or } 1-y=0\\ x=0,y=1\]
anonymous
  • anonymous
ok im so lost lol sorry
anonymous
  • anonymous
you have the same kind of question here, but instead of \(x\) and \(y\) you have \(\cos(x)\) and \(\tan(x)\)
anonymous
  • anonymous
maybe because you are hung up on the trig? lets forget trig for a moment
anonymous
  • 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
  • anonymous
yeah thats probably the problem, im learning though lol
anonymous
  • 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
  • anonymous
yes
anonymous
  • 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
  • anonymous
\[\cos (x)-\cos (x)\tan (x)\]
anonymous
  • anonymous
that is the expression BEFORE it is factored factor out the common factor of cosine then what does it look like ?
anonymous
  • anonymous
\[x-xy=0\]
anonymous
  • anonymous
\[x(1-y)=0\]
anonymous
  • anonymous
\[x=0or 1-y=0\]
anonymous
  • anonymous
ok that was my example with x and y how about with cosine and tangent?
anonymous
  • anonymous
x=0,y=1
anonymous
  • anonymous
ok now we have it with x and y repeat with cosine instead of x and tangent instead of y
anonymous
  • anonymous
both are zero?
anonymous
  • 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
  • anonymous
x=-1?
anonymous
  • anonymous
lets go slow
anonymous
  • anonymous
do you see how i arrived at \[\cos(x)(1-\tan(x))=0\]?
anonymous
  • anonymous
no, i don't know how to do any of this sadly
anonymous
  • anonymous
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anonymous
  • anonymous
opps didnt meant to put that
anonymous
  • anonymous
i am sure you didn't the first step is algebra, not trig
anonymous
  • anonymous
to find X you have to eliminate all the other factors in the equation so that X stands alone
anonymous
  • 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
  • anonymous
do you know the values of \(x\) for which \[cos(x)=0\]?
anonymous
  • anonymous
pi?
anonymous
  • anonymous
not to be mean, but did you just guess?
anonymous
  • anonymous
no, it is not \(\pi\) \[\cos(\pi)=-1\] not zero
anonymous
  • 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
  • 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
anonymous
  • 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
  • anonymous
\[\frac{ \pi }{ 2}\]
anonymous
  • anonymous
yes!!
anonymous
  • anonymous
that is one of them, there is one more
anonymous
  • anonymous
\[\frac{ 3\pi }{ 2}\]
anonymous
  • anonymous
bingo
anonymous
  • anonymous
yay lol
anonymous
  • anonymous
that means \[\cos(\frac{\pi}{2})=0\]and \[\cos(\frac{3\pi}{2})=0\]so those are two of your four answers
anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
now on to tangent we need to solve \[1-\tan(x)=0\]i.e. \[\tan(x)=1\]
anonymous
  • anonymous
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anonymous
  • anonymous
please don't post possible answers we will find them
anonymous
  • anonymous
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anonymous
  • anonymous
we already have two of the four \[\frac{\pi}{2},\frac{3\pi}{2}\]
anonymous
  • anonymous
OH im just posting in general cuz my school is crazy and sometimes the REAL answer is an option
anonymous
  • anonymous
next we need to solve \[\tan(x)=1\] any ideas? (no is a fine answer)
anonymous
  • anonymous
\[\frac{ \pi }{ 4 } and \frac{ 5\pi }{ 4 }\]
anonymous
  • anonymous
yes, not sure how you got that, but it is right
anonymous
  • anonymous
my brother helped me find that awhile ago, i gave up on this question weeks ago ha,
anonymous
  • anonymous
final answer to your question is four numbers \[\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{4},\frac{5\pi}{4}\]
anonymous
  • 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
  • anonymous
since \[\tan(x)=\frac{\sin(x)}{\cos(x)}\] tangent will be one if sine and cosine are equal
anonymous
  • anonymous
yay, thank you so much for helping me figure out the final answer!
anonymous
  • anonymous
yw
anonymous
  • anonymous
C is the correct answer?
anonymous
  • anonymous
\[\frac{ \pi }{ 4} and \frac{ 5\pi }{ 4}\]

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