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burhan101

  • 2 years ago

Solve without a calculator

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

  2. burhan101
    • 2 years ago
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    0<x<2pi

  3. burhan101
    • 2 years ago
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    @UnkleRhaukus

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

  5. burhan101
    • 2 years ago
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    yup

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

  7. UnkleRhaukus
    • 2 years ago
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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

  8. UnkleRhaukus
    • 2 years ago
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    tell me if you get stuck.

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

  10. burhan101
    • 2 years ago
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    what do i do next

  11. UnkleRhaukus
    • 2 years ago
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    multiply by cos^2

  12. burhan101
    • 2 years ago
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    to subtract the two ??

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

  14. UnkleRhaukus
    • 2 years ago
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    how did that happen?

  15. burhan101
    • 2 years ago
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    idk im guessing :S

  16. UnkleRhaukus
    • 2 years ago
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    start again and follow the •'s

  17. burhan101
    • 2 years ago
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    okay

  18. burhan101
    • 2 years ago
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    i'm going to be honest, im confused

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

  20. Hero
    • 2 years ago
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    May I suggest another approach?

  21. UnkleRhaukus
    • 2 years ago
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    ...

  22. burhan101
    • 2 years ago
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    my teacher has a completely different way im confuseeed

  23. UnkleRhaukus
    • 2 years ago
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    well im sure there are many different ways ,

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

  26. UnkleRhaukus
    • 2 years ago
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    i dont like that way as much as i like the way i have outlined , but that is opinion

  27. Yahoo!
    • 2 years ago
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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)

  28. burhan101
    • 2 years ago
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    i am only allowed to have cos and sin in my answers

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

  30. burhan101
    • 2 years ago
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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 ! :$

  31. Hero
    • 2 years ago
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    |dw:1355392376220:dw|

  32. burhan101
    • 2 years ago
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    guys I GOT IT ! :D

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

  34. Hero
    • 2 years ago
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    And what exactly was your teacher's method?

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

  36. burhan101
    • 2 years ago
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    so i just used basic trig and then looking at the unit circle i did elimination and listed the factors

  37. Hero
    • 2 years ago
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    That's not what you posted at the beginning

  38. burhan101
    • 2 years ago
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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

  39. burhan101
    • 2 years ago
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    thats what i did, using the zero principle :

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

  41. burhan101
    • 2 years ago
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    |dw:1355393054138:dw|

  42. Hero
    • 2 years ago
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    Kill your teacher bro

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

  44. Hero
    • 2 years ago
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    @UnkleRhaukus, you ignore negative values?

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

  46. UsukiDoll
    • 2 years ago
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    I'm not saying it sucks or anything though

  47. RolyPoly
    • 2 years ago
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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.

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

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

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

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is replying to Can someone tell me what button the professor is hitting...

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