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Solving a trigonometric equation

  1. Nov 3, 2008 #1
    I'm trying to solve:

    sin2x = -cos2x

    so my first step was...

    2sinxcosx = -cos^2x - sin^2x

    and then...

    2sinxcosx + cos^2x + sin2x = 0

    but I'm confused on how I would end up factoring 2sinxcosx + cos^2x + sin2x, or if I did something wrong earlier in the problem...thank you
  2. jcsd
  3. Nov 3, 2008 #2


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    Can you find another trigonometric identity which would allow you to express the original equation in terms of either sine, or cosine, but not both? Maybe a different double angle identity?
  4. Nov 4, 2008 #3


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    Why not just solve siny= -cos y first and then find x?
  5. Nov 4, 2008 #4
    I'm bad with identities, but wouldn't that be 3*pi/4?

    square root of two over two equals negative negative of the same.
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