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Solving Trigonometric Equations?

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve sin^2(x) = cos^2(x) over the interval [0,2pi]

    3. The attempt at a solution
    sin^2(x) = cos^2(x)
    sin^2(x) = 1-sin^2(x)
    2sin^2(x) = 1
    sin^2(x) = 1/2 = pi/6
    sin(x) = sq root 1/2 = sq root pi/6

    Unfortunately I am completely lost at this point and am not sure if I did that part correctly. If I did where would I go from here? Any help would be greatly appreciated, including any books where I could find some examples of whatever (or what) this is. Regardless, thank you for your time.
  2. jcsd
  3. Apr 21, 2009 #2


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    Where did pi/6 come from in your fourth equation?
  4. Apr 21, 2009 #3
    Sin(1/2) = pi/6
    After I got lost I started adding things to see if I could make my own rules naturally. It has not seemed to work yet though.

    Sorry figured out next few steps
    sin = sq rt 1/2 = 1 sq rt 2 = sq rt 2/2 = pi/4

    Thank you for helping me :).
    Last edited: Apr 21, 2009
  5. Apr 21, 2009 #4


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    You got sin²(x) = ½. Now you have to figure out which x's in [0,2pi] satisfy this equation. One solution is pi/4. There are three more solutions.

    Hint: Draw a picture.
  6. Apr 21, 2009 #5
    pi/4, 3pi/4, 5pi/4, 7pi/4

    I appreciate the help.
  7. Apr 21, 2009 #6
    I am sorry to be a nuisance but I cannot seem to get this.
    The problem statement, all variables and given/known data
    sin(pi(x) + pi/3) = 1
    over interval [-1/3,2/3]

    sin(pi(x))cos(pi/3) + sin(pi/3)cos(pi(x))
    1. 0(1/2) + sq rt 3 /2 - 1 = 1
    - sq rt 3 / 2 = 1

    2. sin(2)(1/2) + sq rt 3 /2 cos(2)

    3. Throw Things -> Cry
  8. Apr 21, 2009 #7


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    I nearly resorted to step 3 as well :cry: until I had a revelation.

    if sin(x)=1
    then x=pi/2,5pi/2,-3pi/2 etc. = pi/2+2pi(n) , n being any integer

    Therefore, sin(pi(x)+pi/3)=1


    I'm sure you can take it from here :smile:
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