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burhan101

Solve without a calculator

  • one year ago
  • one year ago

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

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

    • one year ago
  3. burhan101
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    @UnkleRhaukus

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

    • one year ago
  5. burhan101
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    yup

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

    • one year ago
  7. UnkleRhaukus
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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
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    tell me if you get stuck.

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

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

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

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

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

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

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

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

    • one year ago
  17. burhan101
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    okay

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

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

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

    • one year ago
  21. UnkleRhaukus
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    ...

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

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

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

    • one year ago
  26. UnkleRhaukus
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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!
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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
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    i am only allowed to have cos and sin in my answers

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

    • one year ago
  30. burhan101
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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
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    |dw:1355392376220:dw|

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

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

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

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

    • one year ago
  36. burhan101
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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
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    That's not what you posted at the beginning

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

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

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

    • one year ago
  42. Hero
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    Kill your teacher bro

    • one year ago
  43. UnkleRhaukus
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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
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    @UnkleRhaukus, you ignore negative values?

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

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

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

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

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

    • one year ago
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