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This trig equation is driving me nuts!

  1. Sep 7, 2014 #1
    I have probably put around two hours in to this question to no avail!

    6sin2(x) - 3sin2(2x) + cos2(x)=0

    I have too many fruitless attempts to bother typing them all out.. But my instincts at first told me that this looks like a quadratic equation.

    I have tried using the double angle identity on the 2nd term to get:

    6sin2(x) - 12sin2(x)cos2(x) + cos2(x) = 0 but beyond that I am just messing with identities with no solution in sight.

    Also merging the first and third terms to get :

    5sin2(x) -12sin2(x)cos2(x) + 1 =0

    And much more..

    Please help!

    The answer according to my textbook is +-35.26° +-n360°, +-30° +-n360°
     
    Last edited: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2

    Dick

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    Science Advisor
    Homework Helper

    Try using the identity $$cos(x)^2=1-sin(x)^2$$ to put it in terms of a single trig function. It's actually a quartic. Then factor.
     
  4. Sep 7, 2014 #3
    Thanks I got it now. I had been down that route but for some reason I just never took it all the way..

    Here is the solution I got, though it probably isn't the best.

    6sin2x-3sin22x+cos2x=0

    using Pythagoras and double angle identities

    5sin2x - 12sin2xcos2x + 1 = 0

    sin2x(5 - 12cos2x ) + 1 = 0

    (1 - cos2x)(5 - 12cos2x) + 1 = 0

    12cos4x - 17cos2x + 6 = 0

    (4cos2x - 3)(3cos2x - 2) = 0


    I don't know how I didn't see it before!
     
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