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anonymous

  • 5 years ago

Prove the identity. csc(X)-sin(x)/csc(x)+sin(x)=cos^2x/1+sin^2x

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  1. Yuki
    • 5 years ago
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    sorry I tried, but it doesn't seem straight forward.

  2. anonymous
    • 5 years ago
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    its okay, at least you put effort into it

  3. anonymous
    • 5 years ago
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    This isn't as hard as it looks :)

  4. anonymous
    • 5 years ago
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    \[\csc x =\frac{1}{\sin x}\]by definition, so\[\frac{\csc x - \sin x}{\csc x + \sin x}=\frac{\frac{1}{\sin x}-\sin x}{\frac{1}{\sin x}+\sin x}=\frac{\frac{1-\sin^2x}{\sin x}}{\frac{1+\sin^2 x}{\sin x}}=\frac{1-\sin^2 x}{\sin x}\frac{\sin x}{1+\sin^2x}\]\[=\frac{1-\sin^2 x}{1+\sin^2x}=\frac{\cos^2x}{1+\sin^2x}\]

  5. anonymous
    • 5 years ago
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    Do you need it to be explained more?

  6. anonymous
    • 5 years ago
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    yes if you don't care to

  7. anonymous
    • 5 years ago
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    Is there a specific part?

  8. anonymous
    • 5 years ago
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  9. anonymous
    • 5 years ago
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    You can refer to the part by the number on the attachment, if that's easier.

  10. anonymous
    • 5 years ago
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    brb

  11. anonymous
    • 5 years ago
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    yes that made it alot easier to see thank you! do u care to help me on another one?

  12. anonymous
    • 5 years ago
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    Oh, so you're okay with the last one?

  13. anonymous
    • 5 years ago
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    Yes i knew how to get it started i just didnt know how to simplify that complex fraction, but i understand now.

  14. anonymous
    • 5 years ago
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    Good. What's your other question?

  15. anonymous
    • 5 years ago
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    solve this trig equation on the interval 0<x<2pi, express answer for angles as exact value in radians. 2cos^2x+3cosx+1=0

  16. anonymous
    • 5 years ago
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    Alright...this is a quadratic equation in cos(x). If you 'see' cos(x) as something like, 'u', then you have the equation\[2u^2+3u+1=0\]

  17. amistre64
    • 5 years ago
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    2 or 3 more and we can start a club ;)

  18. anonymous
    • 5 years ago
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    It can be factored into \[(2u+1)(u+1)=0\]

  19. anonymous
    • 5 years ago
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    Yeah, but in 30s you'll be on infinity hero, so... :)

  20. amistre64
    • 5 years ago
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    lol.... by then ill just need another account :)

  21. anonymous
    • 5 years ago
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    No, by then it's like TRON and you're sucked into the website.

  22. anonymous
    • 5 years ago
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    Sorry mj

  23. anonymous
    • 5 years ago
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    is it (2cos^2+1)(cosx+1) btw no harm here i aprreciate your help!

  24. anonymous
    • 5 years ago
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    Yes it is.

  25. anonymous
    • 5 years ago
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    That whole thing is equal to zero, though.

  26. anonymous
    • 5 years ago
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    Which means either 2cos^2x+1 = 0 OR cos(x) + 1 = 0 OR they are BOTH zero. The solution set is all x's that satisfy those conditions (just like a 'normal' quadratic).

  27. anonymous
    • 5 years ago
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    how do i find the angles? thats where i have a problem

  28. anonymous
    • 5 years ago
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    Yep...I'll do them on paper first.

  29. anonymous
    • 5 years ago
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    Okay thanks!

  30. anonymous
    • 5 years ago
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    Okay, if you look at the first factor, you should see something fishy...\[2\cos^2 x + 1 =0 \rightarrow \cos^2x=-\frac{1}{2}\]We have something squared on the left-hand side, and a negative on the right-hand side. If you're only using real numbers (which is what's going on here), this can't happen, so there are no x-values that satisfy this particular equation.

  31. anonymous
    • 5 years ago
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    The square of any real number is always 0 or positive.

  32. anonymous
    • 5 years ago
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    For the second factor, you have\[\cos x = -1\]On the interval \[0 \le x \lt 2\pi\]this happens for \[x=\pi\]only.

  33. anonymous
    • 5 years ago
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    If you plu x=pi into your original equation, you will get 0.

  34. anonymous
    • 5 years ago
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    *plug

  35. anonymous
    • 5 years ago
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    Thanks! if i had (sin(x)-4)(sin(x)+1) how would i find the angles on the same interval?

  36. anonymous
    • 5 years ago
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    Is that equal to zero?

  37. anonymous
    • 5 years ago
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    yes sorry i forgot

  38. anonymous
    • 5 years ago
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    Okay...it's the same deal...always... Either one or both of the factors have to be zero, so you set each one to zero and solve for any possible x-values. Here, in your first factor, you have sin(x)-4 = 0 --> sin(x)=4 This cannot happen (sin(x) oscillates between -1 and 1...there are no values of x here that will punch out a 4, so you won't be getting any solutions from this factor. Note, if it had been something like 4sin(x)-4 = 0 --> sin(x) =4/4 = 1 and you would have solutions). Sot the second factor, you have sin(x)+1=0 which means sin(x) = -1 When does sin(x) = -1 on this interval?

  39. anonymous
    • 5 years ago
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    *'Sot' = 'For'.

  40. anonymous
    • 5 years ago
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    is it -pi/2 +2kpi =3pi/2

  41. anonymous
    • 5 years ago
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    Yes it is.

  42. anonymous
    • 5 years ago
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    3pi/2

  43. anonymous
    • 5 years ago
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    Be careful when using formulas like that; you need to ensure you give the angle that lies in the interval they want (even though -pi/2 is the same as 3pi/2 when considered in the context of plugging into sine, cosine, etc.).

  44. anonymous
    • 5 years ago
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    so sinx=3pi/2

  45. anonymous
    • 5 years ago
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    No, \[\sin \frac{3\pi}{2}=-1\]

  46. anonymous
    • 5 years ago
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    You're trying to find the x's that, when plugged into sine, give -1.

  47. anonymous
    • 5 years ago
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    oh so x=3pi/2 i made a mistake

  48. anonymous
    • 5 years ago
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    Yes.

  49. anonymous
    • 5 years ago
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    you have been a lifesaver, that helped me out alot, especially with the identities.

  50. anonymous
    • 5 years ago
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    You're welcome, mj :)

  51. anonymous
    • 5 years ago
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    So that's it? You're done?

  52. anonymous
    • 5 years ago
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    Uhm this one seems simple but i forgot the identity to use 5sin^2x+2cos^2x=5-3cos^2x

  53. anonymous
    • 5 years ago
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    Okay...this might be more than you have to do, but technically, this is how you would prove it 'properly'.

  54. anonymous
    • 5 years ago
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    Assume that the statement is not true. Then,\[5\sin^2 x + 2\cos^2 x \ne 5-3 \cos^2 x\]But then adding 3cos^2x to both sides gives,\[5\sin^2 x + 5 \cos^2 x \ne 5\]Now, the left-hand side is just\[5\sin^2 x + 5 \cos^2 x=5(\sin^2 x + \cos^2 x) = 5\]which would mean,\[5=5\sin^2 x + 5 \cos^2 x \ne 5\]That is,\[5 \ne 5\]which is false.

  55. anonymous
    • 5 years ago
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    Our assumption that the original statement was false led to a contradiction, so we must take the original statement as true.

  56. anonymous
    • 5 years ago
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    The statement is either true or false. If it's not false, it must me true.

  57. anonymous
    • 5 years ago
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    If you're not happy with that way, I have another.

  58. anonymous
    • 5 years ago
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    i dont think thats what im lookin for my teacher makes us make one side the same as the other side :/

  59. anonymous
    • 5 years ago
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    Yeah...I'll do it another way.

  60. anonymous
    • 5 years ago
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    Take the right-hand side first. You have\[5-3\cos^2 x\]which equals,\[5(1)-3\cos^2 x\]\[=5(\sin^2x + \cos^2x)-3\cos^2x\]\[=5\sin^2x + 5\cos^2x-3\cos^2 x\]\[=5\sin^2 x+2\cos^2x\]

  61. anonymous
    • 5 years ago
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    ahh thats exactly what i was looking for i was thinking you would use sin^2+cos^2=1 for this

  62. anonymous
    • 5 years ago
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    No more questions then?

  63. anonymous
    • 5 years ago
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    tan(x/2)sinx=1-cosx LHS tan(x/2)sinx=1-cos/sin(sin/1)=1-cosx did i do that right?

  64. anonymous
    • 5 years ago
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    I'm not sure what you did. This is what I would do:

  65. anonymous
    • 5 years ago
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    \[\tan \frac{x}{2}\sin x = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}.2\sin \frac{x}{2}\cos \frac{x}{2}=2\sin^2 \frac{x}{2}=1-\cos x\]

  66. anonymous
    • 5 years ago
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    Since by the double angle formula for cosine,\[\cos x = \cos ^2 \frac{x}{2}-\sin^2 \frac{x}{2}=1-2\sin^2 \frac{x}{2} \rightarrow 2\sin^2 \frac{x}{2}=1-\cos x\]

  67. anonymous
    • 5 years ago
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    cos(alpha+beta)/sin(alpha)cos(beta)=cot(alpha)-tan(beta) cos(alpha)cos(beta)-sin(alpha)sin(beta)/.5(sin(alpha+beta)+sin(alpha-beta) how would i finish that

  68. anonymous
    • 5 years ago
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    \[\frac{\cos(\alpha + \beta)}{\sin \alpha \cos \beta}=\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha \cos \beta}{\sin \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha}{\sin \alpha}-\frac{\sin \beta }{\cos \beta}\]\[=\cot \alpha - \tan \beta\]

  69. anonymous
    • 5 years ago
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    That was a much easier way of doing it, identities must take alot of practice to get use to.

  70. anonymous
    • 5 years ago
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    Yeah. It's mostly intuition as to how to start and how to close. That comes from practice!

  71. anonymous
    • 5 years ago
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    I can tell the preparation and follow through is the hardest part! cos(x/2) cos(x/2)=square root of 1+4/5/2 =square root of 9/5/2 =square root of 9/10 is that right if cosx=4/5 with 3pi/2<x<2pi ?

  72. anonymous
    • 5 years ago
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    Um, it's hard for me to tell since I'm not sure of your order of operations.

  73. anonymous
    • 5 years ago
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    What's under the square root?

  74. anonymous
    • 5 years ago
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    \[\frac{\sqrt{1+4/5}}{2}\]?

  75. anonymous
    • 5 years ago
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    i used the half angle formula for cos YES that is correct

  76. anonymous
    • 5 years ago
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    Yes, and it's positive root if you're in that interval.

  77. anonymous
    • 5 years ago
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    yes that is what i used how bout tan(2x) on same interval if i typed out what i have it would be hard for you to understand

  78. anonymous
    • 5 years ago
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    Do you know how to use the equation editor? The button below will take you there.

  79. anonymous
    • 5 years ago
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    oh i didnt know that thank you!

  80. anonymous
    • 5 years ago
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    \[\tan2(\theta)=1-\cos(2\theta)\div2\div1+\cos(2\theta)\div2 \]

  81. anonymous
    • 5 years ago
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    lol...it's not clear...is it...\[\tan 2 \theta =\frac{\frac{1-\cos 2 \theta}{2}}{\frac{1+ \cos 2 \theta}{2}}\]?

  82. anonymous
    • 5 years ago
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    yes thats what i was trying to do i got -3/13 for the answer but im not sure about it

  83. anonymous
    • 5 years ago
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    Are you trying to find tan(2 theta) and the right-hand side is what you came up with?

  84. anonymous
    • 5 years ago
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    yes im trin to find tan(2theta) on the interval 3pi/2<x<2pi i used the double angle formula for tan

  85. anonymous
    • 5 years ago
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    For you to get a number, tan(2theta) would have to have been set equal to something in the first place. Is that what you have?

  86. anonymous
    • 5 years ago
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    Because tan(2theta) takes an infinity of values on that interval...unless tan(2theta)=(something) to begin with.

  87. anonymous
    • 5 years ago
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    yes thats what i have it equal to

  88. anonymous
    • 5 years ago
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    ok...let me see.

  89. anonymous
    • 5 years ago
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    okay!

  90. anonymous
    • 5 years ago
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    I'll tell you what I'm doing (because I'm being distracted)...since cos2theta is all the rage in this equation, try to get everything in terms of cos(2theta) and determine a polynomial in cos(2theta),\[P(\cos 2 \theta)=0\]Then solve for the roots of the polynomial.

  91. anonymous
    • 5 years ago
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    okay i will do that! can you tell me how to simplify real quick (-1/2)(4/5)-square root of 3/2(3/5)

  92. anonymous
    • 5 years ago
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    I need you to use the equation editor ::

  93. anonymous
    • 5 years ago
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    \[-1/2(4/5)-(\sqrt{3/2})(3/5)\]

  94. anonymous
    • 5 years ago
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    \[-\frac{1}{2}\frac{4}{5}-\frac{3}{5}\sqrt{\frac{3}{2}}\]do you mean this?

  95. anonymous
    • 5 years ago
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    yess that is right

  96. anonymous
    • 5 years ago
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  97. anonymous
    • 5 years ago
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    Thank you for everything! I appreciate your time!

  98. anonymous
    • 5 years ago
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    welcome

  99. anonymous
    • 5 years ago
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    did you finish the last one?

  100. anonymous
    • 5 years ago
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    nope :/

  101. anonymous
    • 5 years ago
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    Is this due online or something?

  102. anonymous
    • 5 years ago
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    no i have to turn in next week

  103. anonymous
    • 5 years ago
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    Okay. I have to go sort some things out (like my life). I can look at the last problem and post back here. You should get an e-mail from the site saying it's been updated. I'll probably be back during the day to do it...probably :)

  104. anonymous
    • 5 years ago
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    Alright sounds good! I'll be sure to check back on here for it! Have a good day kind sir

  105. anonymous
    • 5 years ago
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    You too ;)

  106. anonymous
    • 5 years ago
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    Here you are...remember to check it.

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  108. anonymous
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  112. anonymous
    • 5 years ago
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    So, in summary, you have a quartic polynomial in tan(theta), and so, four possible roots. Two of these roots will be complex conjugates of each other, so are not included. The roots are:\[\tan \theta = 0\]\[\tan \theta \approx -1.52138\]which means that your angles are (for this interval)\[\theta = 2\pi\](from the first root)\[\theta \approx 5.294^c\](from the second root). The 'c' superscript means, 'radians'.

  113. anonymous
    • 5 years ago
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    The plot is for f(t)=t^3-t+2

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