## anonymous one year ago What values for (posted below) satisfy the equation?

1. anonymous

2. anonymous

3. anonymous

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

4. 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

5. anonymous

you good from there?

6. anonymous

I have no clue how to factor these problems

7. anonymous

i factored it for you

8. anonymous

How do I get to the final answer?

9. 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

10. anonymous

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

11. anonymous

subtract y

12. anonymous

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

13. anonymous

ok im so lost lol sorry

14. anonymous

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

15. anonymous

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

16. 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

17. anonymous

yeah thats probably the problem, im learning though lol

18. 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?

19. anonymous

yes

20. 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)

21. anonymous

$\cos (x)-\cos (x)\tan (x)$

22. anonymous

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

23. anonymous

$x-xy=0$

24. anonymous

$x(1-y)=0$

25. anonymous

$x=0or 1-y=0$

26. anonymous

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

27. anonymous

x=0,y=1

28. anonymous

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

29. anonymous

both are zero?

30. anonymous

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

31. anonymous

x=-1?

32. anonymous

lets go slow

33. anonymous

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

34. anonymous

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

35. anonymous

36. anonymous

opps didnt meant to put that

37. anonymous

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

38. anonymous

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

39. 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 $1-\tan(x)=0$

40. anonymous

do you know the values of $$x$$ for which $cos(x)=0$?

41. anonymous

pi?

42. anonymous

not to be mean, but did you just guess?

43. anonymous

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

44. 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

45. 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

46. 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

47. anonymous

$\frac{ \pi }{ 2}$

48. anonymous

yes!!

49. anonymous

that is one of them, there is one more

50. anonymous

$\frac{ 3\pi }{ 2}$

51. anonymous

bingo

52. anonymous

yay lol

53. anonymous

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

54. anonymous

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56. anonymous

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

57. anonymous

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60. anonymous

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

61. anonymous

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

62. anonymous

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

63. anonymous

$\frac{ \pi }{ 4 } and \frac{ 5\pi }{ 4 }$

64. anonymous

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

65. anonymous

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

66. anonymous

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

67. 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

68. anonymous

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

69. anonymous

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

70. anonymous

yw

71. anonymous

$\frac{ \pi }{ 4} and \frac{ 5\pi }{ 4}$