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anonymous

  • one year ago

What values for (posted below) satisfy the equation?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    the first attachment fills in the blank, the second one follows the sentence

  4. anonymous
    • one year ago
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    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

  5. anonymous
    • one year ago
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    you good from there?

  6. anonymous
    • one year ago
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    I have no clue how to factor these problems

  7. anonymous
    • one year ago
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    i factored it for you

  8. anonymous
    • one year ago
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    How do I get to the final answer?

  9. anonymous
    • one year ago
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    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

  10. anonymous
    • one year ago
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    suppose the question said "solve for \(x\) and \(y\) \(x-xy=0\)" what would you do?

  11. anonymous
    • one year ago
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    subtract y

  12. anonymous
    • one year ago
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    hmm no \[x-xy=0\\ x(1-y)=0\\ x=0\text {or } 1-y=0\\ x=0,y=1\]

  13. anonymous
    • one year ago
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    ok im so lost lol sorry

  14. anonymous
    • one year ago
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    you have the same kind of question here, but instead of \(x\) and \(y\) you have \(\cos(x)\) and \(\tan(x)\)

  15. anonymous
    • one year ago
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    maybe because you are hung up on the trig? lets forget trig for a moment

  16. anonymous
    • one year ago
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    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

  17. anonymous
    • one year ago
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    yeah thats probably the problem, im learning though lol

  18. anonymous
    • one year ago
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    so we start by factoring the expression \[\cos(x)-\cos(x)\tan(x)\] is it clear that each term has a cosine in it?

  19. anonymous
    • one year ago
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    yes

  20. anonymous
    • one year ago
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    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)

  21. anonymous
    • one year ago
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    \[\cos (x)-\cos (x)\tan (x)\]

  22. anonymous
    • one year ago
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    that is the expression BEFORE it is factored factor out the common factor of cosine then what does it look like ?

  23. anonymous
    • one year ago
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    \[x-xy=0\]

  24. anonymous
    • one year ago
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    \[x(1-y)=0\]

  25. anonymous
    • one year ago
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    \[x=0or 1-y=0\]

  26. anonymous
    • one year ago
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    ok that was my example with x and y how about with cosine and tangent?

  27. anonymous
    • one year ago
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    x=0,y=1

  28. anonymous
    • one year ago
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    ok now we have it with x and y repeat with cosine instead of x and tangent instead of y

  29. anonymous
    • one year ago
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    both are zero?

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

  31. anonymous
    • one year ago
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    x=-1?

  32. anonymous
    • one year ago
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    lets go slow

  33. anonymous
    • one year ago
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    do you see how i arrived at \[\cos(x)(1-\tan(x))=0\]?

  34. anonymous
    • one year ago
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    no, i don't know how to do any of this sadly

  35. anonymous
    • one year ago
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  36. anonymous
    • one year ago
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    opps didnt meant to put that

  37. anonymous
    • one year ago
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    i am sure you didn't the first step is algebra, not trig

  38. anonymous
    • one year ago
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    to find X you have to eliminate all the other factors in the equation so that X stands alone

  39. anonymous
    • one year ago
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    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 \[1-\tan(x)=0\]

  40. anonymous
    • one year ago
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    do you know the values of \(x\) for which \[cos(x)=0\]?

  41. anonymous
    • one year ago
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    pi?

  42. anonymous
    • one year ago
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    not to be mean, but did you just guess?

  43. anonymous
    • one year ago
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    no, it is not \(\pi\) \[\cos(\pi)=-1\] not zero

  44. anonymous
    • one year ago
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    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

  45. anonymous
    • one year ago
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    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

  46. anonymous
    • one year ago
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    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

  47. anonymous
    • one year ago
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    \[\frac{ \pi }{ 2}\]

  48. anonymous
    • one year ago
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    yes!!

  49. anonymous
    • one year ago
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    that is one of them, there is one more

  50. anonymous
    • one year ago
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    \[\frac{ 3\pi }{ 2}\]

  51. anonymous
    • one year ago
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    bingo

  52. anonymous
    • one year ago
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    yay lol

  53. anonymous
    • one year ago
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    that means \[\cos(\frac{\pi}{2})=0\]and \[\cos(\frac{3\pi}{2})=0\]so those are two of your four answers

  54. anonymous
    • one year ago
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  55. anonymous
    • one year ago
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  56. anonymous
    • one year ago
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    now on to tangent we need to solve \[1-\tan(x)=0\]i.e. \[\tan(x)=1\]

  57. anonymous
    • one year ago
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  58. anonymous
    • one year ago
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    please don't post possible answers we will find them

  59. anonymous
    • one year ago
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  60. anonymous
    • one year ago
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    we already have two of the four \[\frac{\pi}{2},\frac{3\pi}{2}\]

  61. anonymous
    • one year ago
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    OH im just posting in general cuz my school is crazy and sometimes the REAL answer is an option

  62. anonymous
    • one year ago
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    next we need to solve \[\tan(x)=1\] any ideas? (no is a fine answer)

  63. anonymous
    • one year ago
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    \[\frac{ \pi }{ 4 } and \frac{ 5\pi }{ 4 }\]

  64. anonymous
    • one year ago
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    yes, not sure how you got that, but it is right

  65. anonymous
    • one year ago
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    my brother helped me find that awhile ago, i gave up on this question weeks ago ha,

  66. anonymous
    • one year ago
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    final answer to your question is four numbers \[\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{4},\frac{5\pi}{4}\]

  67. anonymous
    • one year ago
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    \[\frac{\pi}{4}\]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

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

  69. anonymous
    • one year ago
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    yay, thank you so much for helping me figure out the final answer!

  70. anonymous
    • one year ago
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    yw

  71. anonymous
    • one year ago
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    C is the correct answer?

  72. anonymous
    • one year ago
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    \[\frac{ \pi }{ 4} and \frac{ 5\pi }{ 4}\]

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