1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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.


    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted