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britbrat4290 Group Title

Solve the equation on the interval [0, 2π). sin2 x - cos2 x = 0

  • one year ago
  • one year ago

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  1. britbrat4290 Group Title
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    sry no lol

    • one year ago
  2. Coolsector Group Title
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    and you should convince yourself why it is fine to divide by cos2x . ( it is fine. )

    • one year ago
  3. britbrat4290 Group Title
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    wud it b pie/4, pie/6 ?

    • one year ago
  4. Coolsector Group Title
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    wait im stupid

    • one year ago
  5. Coolsector Group Title
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    tan(2x) = 1 !!

    • one year ago
  6. britbrat4290 Group Title
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    ??

    • one year ago
  7. Coolsector Group Title
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    so the general solution for tan(x) = tan(y) x = y + pi*k so in our case: tan(2x) = tan(pi/4) 2x = pi/4 + pi*k x = pi/8 + (pi/2) * k

    • one year ago
  8. britbrat4290 Group Title
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    pie/4 just then

    • one year ago
  9. Coolsector Group Title
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    no

    • one year ago
  10. britbrat4290 Group Title
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    u got over 8?

    • one year ago
  11. Coolsector Group Title
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    pi/8 , 5pi/8 , 9pi/8 , 13pi/8

    • one year ago
  12. Coolsector Group Title
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    i first wrote tan(2x) = 0 which was WRONG

    • one year ago
  13. britbrat4290 Group Title
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    pi/4, 3pi/4, 5pi/4, 7pi/4

    • one year ago
  14. Coolsector Group Title
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    no, x=pi/8 , 5pi/8 , 9pi/8 , 13pi/8

    • one year ago
  15. britbrat4290 Group Title
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    hmm

    • one year ago
  16. Coolsector Group Title
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    wait lol was it :

    • one year ago
  17. Coolsector Group Title
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    \[\sin(2x) - \cos(2x) = 0\]

    • one year ago
  18. Coolsector Group Title
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    or \[\sin^2(x) - \cos^2(x) = 0\]

    • one year ago
  19. Coolsector Group Title
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    cause i solved for the FIRST one

    • one year ago
  20. Coolsector Group Title
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    ?

    • one year ago
  21. britbrat4290 Group Title
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    it said i got it right for the four i picked? weird

    • one year ago
  22. Coolsector Group Title
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    but what was the correct question i wrote here two

    • one year ago
  23. Coolsector Group Title
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    i think i solved for the wrong one

    • one year ago
  24. britbrat4290 Group Title
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    im lost sry its ok bc i got it right so its all good :)

    • one year ago
  25. Coolsector Group Title
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    cause if the question was : sin^2(x) - cos^2(x) = 0 then we get tan^2(x) = 1 tan(x) = +/-1 so 1. tan(x) = 1 tan(x) = tan(pi/4) x = pi/4 + pi * k x = pi/4 , 5pi/4 2.tan(x) = -1 tan(x) = tan(-pi/4) x = -pi/4 + pi*k x = 3pi/4 , 7pi/4 so : x = pi/4, 3pi/4, 5pi/4, 7pi/4

    • one year ago
  26. Coolsector Group Title
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    and i thought it was sin(2x) - cos(2x) = 0

    • one year ago
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