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Hannah_Ahn Group Title

solve algebraically for x: 4sin^2x=3tan^2-1 (answers: x=pi over 4, 3pi over 4, 5pi over 4, and 7pi over 4)

  • 2 years ago
  • 2 years ago

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  1. shankvee\ Group Title
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    is that tan^2 x?

    • 2 years ago
  2. Hannah_Ahn Group Title
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    yes!

    • 2 years ago
  3. shankvee\ Group Title
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    Ok this you can do, Just juggle about with identites first multiply both sides by cos ^2 x, LHS becomes 4sin^2 x cos ^2 x=(sin 2x)^2. Did you understand how?

    • 2 years ago
  4. shankvee\ Group Title
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    Did you understand till here lets go step by step.

    • 2 years ago
  5. Hannah_Ahn Group Title
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    shouldn't it be 2sin1/2^2

    • 2 years ago
  6. shankvee\ Group Title
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    Do you know the identity 2sinx cosx=sin2x

    • 2 years ago
  7. Hannah_Ahn Group Title
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    yep

    • 2 years ago
  8. Hannah_Ahn Group Title
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    2sin2^2*

    • 2 years ago
  9. shankvee\ Group Title
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    good so you have here 2sinx cos x the whole squared which is sin 2x the whole squared.

    • 2 years ago
  10. Hannah_Ahn Group Title
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    2sin2x^2**

    • 2 years ago
  11. shankvee\ Group Title
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    So On the LHS there is sin ^2 2x which we can write it as 1- cos ^2 2x

    • 2 years ago
  12. Hannah_Ahn Group Title
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    shouldn't it be 2sin^2 2x??

    • 2 years ago
  13. shankvee\ Group Title
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    no do it agian 2 wont come.

    • 2 years ago
  14. Hannah_Ahn Group Title
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    4sin^2xcos^2 = 2sin2x^2 not sin2x^2 ?

    • 2 years ago
  15. shankvee\ Group Title
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    no it is square in this one 2sinx cos x becomes sin 2x and another 2sinx cos x becomes sin 2x you need two 2's to make (sin x cosx)^2 (sin 2x)^2

    • 2 years ago
  16. Hannah_Ahn Group Title
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    OHHH!!!! I didn't know that .. :P thanks, let's keep it going

    • 2 years ago
  17. shankvee\ Group Title
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    So yeah write sin ^2 2x as 1- cos ^2 2x. ok?

    • 2 years ago
  18. Hannah_Ahn Group Title
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    kk

    • 2 years ago
  19. shankvee\ Group Title
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    Ok so now come to RHS 3tan ^2 x * cos^2 x - cos ^2 x

    • 2 years ago
  20. shankvee\ Group Title
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    tan ^2 x *cos ^2 x=sin^2 x

    • 2 years ago
  21. shankvee\ Group Title
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    so RHS becomes 3sin^2 x - cos^2 x

    • 2 years ago
  22. shankvee\ Group Title
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    Now do you know the identities sin ^2 x=(1-cos2x)/2 and cos ^2 x=(1+cos 2x)/2

    • 2 years ago
  23. Hannah_Ahn Group Title
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    i don't know abhout the /2

    • 2 years ago
  24. Hannah_Ahn Group Title
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    sin^2x = 1-cos^2x

    • 2 years ago
  25. shankvee\ Group Title
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    no no that identity is \[\sin ^{2}x=(1-\cos 2x)/2\]

    • 2 years ago
  26. Hannah_Ahn Group Title
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    over 2 ?

    • 2 years ago
  27. Hannah_Ahn Group Title
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    isn't it pythagorean indentities

    • 2 years ago
  28. shankvee\ Group Title
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    ITS \[\cos 2x \] not\[\cos ^{2}x\]

    • 2 years ago
  29. Hannah_Ahn Group Title
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    i am so sorry. :S i didn't cover that identities yet

    • 2 years ago
  30. Hannah_Ahn Group Title
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    those*

    • 2 years ago
  31. shankvee\ Group Title
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    You have to use those identities you have 1- cos ^2 2x on the LHS so write everything in the RHS in terms of cos 2x

    • 2 years ago
  32. shankvee\ Group Title
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    even if you didn't know them now you know them... so substitue for sin ^2 x and cos ^2 x by using the identities i told you and tell me what you get...

    • 2 years ago
  33. shankvee\ Group Title
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    did you get what i'm trying to tell?

    • 2 years ago
  34. Hannah_Ahn Group Title
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    yes

    • 2 years ago
  35. shankvee\ Group Title
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    Good so tell me what do you have on the RHS...

    • 2 years ago
  36. Hannah_Ahn Group Title
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    unfortunately, i cannot work it out with these identities that i didn't learn for a test. .

    • 2 years ago
  37. shankvee\ Group Title
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    I'm sorry i don't know how to solve the problem in any other way.... Anyway if you want to continue with my method take LCM and all and simplify both sides you get cos ^2 2x=2cos 2x ths means cos 2x=0 and cos 2x=2(This case is not possible so the answer is just cos 2x=0

    • 2 years ago
  38. Hannah_Ahn Group Title
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    it's okay I still appreciate you for what you've done. :)

    • 2 years ago
  39. shankvee\ Group Title
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    Okay here's an alterative easier solution , You can use sec ^2 x= 1+tan ^2 x

    • 2 years ago
  40. shankvee\ Group Title
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    So in the RHS write tan ^2 x as sec ^2 x -1 and simplify you get 8-4cos^2 x= 3sec^2 x. See if you understand till here....

    • 2 years ago
  41. shankvee\ Group Title
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    Any doubts in what i have done? If not let us say cos^2 x is some t then obviously sec ^2 x=1/cos^2 x=1/t so you get 8-4t=3/t. 8t-4t^2=3. 4t^2-8t+3=0 this implies t=1/2 or t=3/2 t=3/2 is not possible hence t=1/2 or cos ^2 x=1/2 whose obvious solutions are pi/4,3pi/4 and so on.

    • 2 years ago
  42. shankvee\ Group Title
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    lalaly wrong the equation when divided as by you becomes 4=3/cos^2x - cosec^2x

    • 2 years ago
  43. lalaly Group Title
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    oops sorry

    • 2 years ago
  44. shankvee\ Group Title
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    did you get it hannah?

    • 2 years ago
  45. Hannah_Ahn Group Title
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    sorry i took a break . i am keep looking to figure out what you have done .

    • 2 years ago
  46. Hannah_Ahn Group Title
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    i feel like you are one of my tutor that i had in my past years :) haha he was great. anyway ..

    • 2 years ago
  47. shankvee\ Group Title
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    What a compliment \m/

    • 2 years ago
  48. shankvee\ Group Title
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    Do you know how to solve quadratic equations?

    • 2 years ago
  49. Hannah_Ahn Group Title
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    i learned it i will need to review it

    • 2 years ago
  50. shankvee\ Group Title
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    Anyway read what i have done properly its not too hard to understand i've just substitued tan^2 x as sec ^2 x -1

    • 2 years ago
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