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Trigonometric equation

  1. Sep 16, 2013 #1

    GuN

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    1. The problem statement, all variables and given/known data

    2 sin(2x) = 2 cos(x)

    2. Relevant equations



    3. The attempt at a solution

    4sinxcosx-2cosx=0

    2(2sinxcosx-cosx)=0

    2cosx(sinx-1)=0

    2cosx=0 sinx=1


    And then I did the math and got pi/2 and 3pi/2 , but that's apparently wrong.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2013 #2

    jedishrfu

    Staff: Mentor

    I get cosx * (2*sinx - 1) = 0 for your second to last step. Does that help?
     
  4. Sep 16, 2013 #3

    GuN

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    I still get pi/2 and 3 pi/2.

    >_<

    I'll try consulting my notes on how to properly calculate cosx=0 and sinx=1/2.
     
  5. Sep 16, 2013 #4

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    How? You know that ##\sin (\pi/2) = 1## and ##\sin (3\pi/2) = -1##. So obviously neither of those satisfy ##\sin x = 1/2##.
     
  6. Sep 16, 2013 #5
    arcsin (1/2) = x1
    arccos (0) = x2

    You have done everything correctly, now you need to figure out what the values for x are. This is not a linear equation, there are 2 series of solutions.

    The system I showed you will not be your final answer.
     
  7. Sep 17, 2013 #6

    GuN

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    Right, I double checked the notes and found I used the wrong radians.

    I redid it and got (for the domain of [0, 2pi] ) pi/6, 5pi/6, pi/2 and 3pi/2.
     
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