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burhan101 Group Title

Solve without a calculator

  • one year ago
  • one year ago

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  1. burhan101 Group Title
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    \[\[\huge 4\tan^2-\sec^2x=0\]\]

    • one year ago
  2. burhan101 Group Title
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    0<x<2pi

    • one year ago
  3. burhan101 Group Title
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    @UnkleRhaukus

    • one year ago
  4. UnkleRhaukus Group Title
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    \[4\tan^2 (x)-\sec^2 (x)=0\]

    • one year ago
  5. burhan101 Group Title
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    yup

    • one year ago
  6. UnkleRhaukus Group Title
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    \[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\]\[\sec(\theta)=\frac{1}{\cos(\theta)}\]

    • one year ago
  7. UnkleRhaukus Group Title
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    •take the sec term to the other side of the equation, •substitute the above definitions into your equation, •multiply by the denominator •divide by 4 •take the sqrt of both sides , •take the inverse trig function to isolate x

    • one year ago
  8. UnkleRhaukus Group Title
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    tell me if you get stuck.

    • one year ago
  9. burhan101 Group Title
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    \[4\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos^2x }=0\]

    • one year ago
  10. burhan101 Group Title
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    what do i do next

    • one year ago
  11. UnkleRhaukus Group Title
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    multiply by cos^2

    • one year ago
  12. burhan101 Group Title
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    to subtract the two ??

    • one year ago
  13. burhan101 Group Title
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    (2sinx+1)(2sinx-1) / cos^2x

    • one year ago
  14. UnkleRhaukus Group Title
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    how did that happen?

    • one year ago
  15. burhan101 Group Title
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    idk im guessing :S

    • one year ago
  16. UnkleRhaukus Group Title
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    start again and follow the •'s

    • one year ago
  17. burhan101 Group Title
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    okay

    • one year ago
  18. burhan101 Group Title
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    i'm going to be honest, im confused

    • one year ago
  19. burhan101 Group Title
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    by following the checlist this is what im getting sinxcosx-sinx

    • one year ago
  20. Hero Group Title
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    May I suggest another approach?

    • one year ago
  21. UnkleRhaukus Group Title
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    ...

    • one year ago
  22. burhan101 Group Title
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    my teacher has a completely different way im confuseeed

    • one year ago
  23. UnkleRhaukus Group Title
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    well im sure there are many different ways ,

    • one year ago
  24. burhan101 Group Title
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    • one year ago
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  25. Yahoo! Group Title
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    1 + tan^2 x = sec^2 x tan^2 x = sec^2 x - 1

    • one year ago
  26. UnkleRhaukus Group Title
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    i dont like that way as much as i like the way i have outlined , but that is opinion

    • one year ago
  27. Yahoo! Group Title
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    4 ( sec^2 x - 1) - sec^2 x = 0 4 sec^2 x - 4 - sec^2 x = 0 3 sec^2 x = 4 sec x = 2/ sqrt (3)

    • one year ago
  28. burhan101 Group Title
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    i am only allowed to have cos and sin in my answers

    • one year ago
  29. Yahoo! Group Title
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    Lol.....u shuld have Mentioned that in the question...)

    • one year ago
  30. burhan101 Group Title
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    Steps: 1. rearrange and isolate the unknown 2. move everything to one since, set equal to zero 3. use the zero principle 4. use inverse trig to find the angle 5. use CAST or ASTC to adjust answers @Yahoo! i know sorry ! :$

    • one year ago
  31. Hero Group Title
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    |dw:1355392376220:dw|

    • one year ago
  32. burhan101 Group Title
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    guys I GOT IT ! :D

    • one year ago
  33. burhan101 Group Title
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    i just used my teachers method but everyone's explanations helped. thankyou :D

    • one year ago
  34. Hero Group Title
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    And what exactly was your teacher's method?

    • one year ago
  35. burhan101 Group Title
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    we're supposed to find what x can and cannot equal

    • one year ago
  36. burhan101 Group Title
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    so i just used basic trig and then looking at the unit circle i did elimination and listed the factors

    • one year ago
  37. Hero Group Title
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    That's not what you posted at the beginning

    • one year ago
  38. burhan101 Group Title
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    burhan101 0 Steps: 1. rearrange and isolate the unknown 2. move everything to one since, set equal to zero 3. use the zero principle 4. use inverse trig to find the angle 5. use CAST or ASTC to adjust answers

    • one year ago
  39. burhan101 Group Title
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    thats what i did, using the zero principle :

    • one year ago
  40. Hero Group Title
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    You mind showing some of those steps using the drawing button like I did?

    • one year ago
  41. burhan101 Group Title
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    |dw:1355393054138:dw|

    • one year ago
  42. Hero Group Title
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    Kill your teacher bro

    • one year ago
  43. UnkleRhaukus Group Title
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    \[4\tan^2 (x)-\sec^2 (x)=0\]\[4\tan^2 (x)=\sec^2 (x)\]\[4\left(\frac{\sin(x)}{\cos (x)}\right)^2 =\left(\frac1{\cos(x)}\right)^2\]\[4\frac{\sin^2(x)}{\cos^2 (x)} =\left(\frac1{\cos^2(x)}\right)\]\[4\sin^2(x)=\frac{\cos^2(x)}{\cos^2(x)}\]\[4\sin^2(x) =1\]\[\sin^2(x) =\frac14\]\[\sin(x)=\frac 12\]\[x=\arcsin\left(\frac12\right)=60°\]

    • one year ago
  44. Hero Group Title
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    @UnkleRhaukus, you ignore negative values?

    • one year ago
  45. UsukiDoll Group Title
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    wow Hero that's just harsh, but...yeah..no comment

    • one year ago
  46. UsukiDoll Group Title
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    I'm not saying it sucks or anything though

    • one year ago
  47. RolyPoly Group Title
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    \[4\tan^2 (x)-\sec^2 (x)=0\]\[4\tan^2 (x)-(1+tan^2x)=0\]\[3\tan^2 (x)-1=0\]\[tan^2x = \frac{1}{3}\]That shouldn't be difficult.

    • one year ago
  48. RolyPoly Group Title
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    Just use the identity \(1+tan^2x=sec^2x\)

    • one year ago
  49. UnkleRhaukus Group Title
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    @hero 0<x<2pi stated in question

    • one year ago
  50. Hero Group Title
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    I didn't see that second line. He posted it in pieces.

    • one year ago
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is replying to Can someone tell me what button the professor is hitting...

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