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uri

  • 3 years ago

Halp meh pliss, The number of positive values of x less than 360 degrees which satisfy the equation sin 1/2x =cos x is? a)0 b)1) c)2 d)3 e)4

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  1. saifoo.khan
    • 3 years ago
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    You do math too? D:

  2. hba
    • 3 years ago
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    Mujhe bhi yaqeen nahi horaha :P

  3. anonymous
    • 3 years ago
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    looks like there are two http://www.wolframalpha.com/input/?i=sin%28x%2F2%29%2Ccos%28x%29+domain+0..2pi

  4. yrelhan4
    • 3 years ago
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    write cosx in terms of sin^2 x/2 and solve!

  5. uri
    • 3 years ago
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    1/2x=sin^2x/2 Right ?

  6. yrelhan4
    • 3 years ago
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    cosx= 1-2sin^2 (x/2)

  7. yrelhan4
    • 3 years ago
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    you get a quadratic.. solve for x..

  8. uri
    • 3 years ago
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    2sin^2(x/2)+cosx-1=0

  9. yrelhan4
    • 3 years ago
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    2t^2 + t -1=0 where t=sin(x/2)

  10. uri
    • 3 years ago
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    Quad,breaking the middle term or completing the square method?

  11. yrelhan4
    • 3 years ago
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    breaking the middle term.. you would get (2t-1)(t+1)=0 where t=sin(x/2)

  12. uri
    • 3 years ago
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    eh,2t-1=0=)t=1/2 t+1=0=)t=-1

  13. yrelhan4
    • 3 years ago
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    yup!

  14. uri
    • 3 years ago
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    Sin(x/2)=1/2 Sin(x/2)=-1

  15. yrelhan4
    • 3 years ago
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    so there would be 3 values of x/2 in 0 to 360.. 30,120,270. so x=60,240,540.. since you want from 0 to 360.. there would be 2 solns..

  16. uri
    • 3 years ago
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    it was Sin x/2

  17. yrelhan4
    • 3 years ago
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    if sin(x/2) = 1/2 , what is x? 0>x>360

  18. hba
    • 3 years ago
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    Oh so, sin x/2 =cos x ?

  19. yrelhan4
    • 3 years ago
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    1- sin^2(x/2)= cosx @hba

  20. uri
    • 3 years ago
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    em BOREEEEEED.

  21. yrelhan4
    • 3 years ago
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    look.. let x/2=k .. sin(k)=1/2 --> k=30, 120.. --> x=60, 240

  22. yrelhan4
    • 3 years ago
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    cosx= 1-2sin^2 (x/2) @hba missed the two..

  23. yrelhan4
    • 3 years ago
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    thik hai? @uri

  24. sirm3d
    • 3 years ago
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    @yrelhan4 since \(\displaystyle 0 \leq x \leq 360\) then \(\displaystyle 0 \leq \frac{x}{2} \leq 180\)

  25. sirm3d
    • 3 years ago
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    now, \(\sin x/2 = 1/2\) therefore \(x/2=30, 150\) or \(x=60, 300\)

  26. uri
    • 3 years ago
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    Yrelhan GUESSSS WHAT?! I donyt undastand aaaanythang nd ima sho confuzzeld tooh,I caant eveen speel corractly. o.O

  27. yrelhan4
    • 3 years ago
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    lol.. i am such a fool.. you are right! @sirm3d

  28. uri
    • 3 years ago
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    Wut iz the aaansweeer?! none of my option matches your answer!

  29. yrelhan4
    • 3 years ago
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    i just wrotw150 value wrong, right? why am i a fool then? :P

  30. sirm3d
    • 3 years ago
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    the other equation, \(\sin x/2 = -1\) has no solution in \(0 \leq \frac{x}{2} \leq 180\)

  31. yrelhan4
    • 3 years ago
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    @uri 2.. final!

  32. uri
    • 3 years ago
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    @sirm3d sinx/2=1 has no solution so we will leave it? xD

  33. yrelhan4
    • 3 years ago
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    @uri its sin(x/2) = -1

  34. uri
    • 3 years ago
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    k.

  35. sirm3d
    • 3 years ago
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    @uri, just count the number of solutions: \(x=60^0,\;300^0\)

  36. uri
    • 3 years ago
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    @sirm3d 2 solutions?

  37. sirm3d
    • 3 years ago
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    yup. (C) 2

  38. yrelhan4
    • 3 years ago
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    are kitni baar puchegi? @uri :P

  39. uri
    • 3 years ago
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    AMG THANKYOUUUUUUUUUUUUU!AMG THANKYOUU SOO MUCHHHHH!!!! :D

  40. yrelhan4
    • 3 years ago
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    mera medal.. :'( .. aadha to maine solve kiya..

  41. yrelhan4
    • 3 years ago
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    huh!

  42. uri
    • 3 years ago
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    You have 2 now :P

  43. yrelhan4
    • 3 years ago
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    haha.. he's got 2 now too!

  44. uri
    • 3 years ago
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    Happy? :D

  45. yrelhan4
    • 3 years ago
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    ha! yes.. :P

  46. uri
    • 3 years ago
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    Yw xD

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