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    Solve sin4θ = cos2θ

    What do you do with the sin4θ? I tried turning it into 2sin2θcos2θ using the double angle formulae but I don't know where to go from there...

    Because say if I then did:

    2(2sinθcosθ)(1-2sin2θ) = 1-2sin2θ

    or...

    2(2sinθcosθ)(2cos2θ-1) = wtf shall I use




    I don't even know what to do- am I supposed to make everything so it's all in terms of sinθ or cosθ?? Nothing is seeming to work...
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    Absolute first thing you do: set \alpha = 2\theta, so your equation becomes \sin 2\alpha = \cos \alpha. Solve for alpha, and then divide by 2 to find theta.
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    (Original post by DFranklin)
    Absolute first thing you do: set \alpha = 2\theta, so your equation becomes \sin 2\alpha = \cos \alpha. Solve for alpha, and then divide by 2 to find theta.
    Where would you get α from?
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    (Original post by jessyjellytot14)
    Where would you get α from?
    It's just a substitution for the sake of simplicity, you can choose \lambda = 2\theta if you want, doesn't matter.

    However, I do not see why you are expressing \cos(2\theta) in a different form. Simply get everything on one side at 2\sin(2\theta)\cos(2\theta) stage, and just factor out a \cos(2\theta). You can then solve for theta.
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    (Original post by RDKGames)
    Simply get everything on one side at 2\sin(2\theta)\cos(2\theta) stage, and just factor out a \cos(2\theta). You can then solve for theta.
    Oooooh, that is the logical thing to do! How did I not think of that? :facepalm:
 
 
 
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