tan^2x=sin2x, 0<=x<2pi
cosx cannot = 0 ; x cannot = pi/2 or 3pi/2
sin^2x/cos^2x=2sinxcosx
sin^2x=2sinxcos^3x
sin^2x-2sinxcos^3x=0
sinx(sinx-2cos^3x)=0
sinx=0 or sinx-2cos^3x=0
+-sqrt(1-cos^2x) - 2cos^3x=0
+-sqrt(1-cos^2x) = 2cos^3x
*square both sides to remove radical
1-cos^2x = 4cos^6x
4cos^6x-cos^2x-1=0
I don't know how to solve this equation, can someone please provide me with a link to read how to solve this equiation?
First of all, don't forget that you need to solveOriginally Posted by ostheimerdl
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Anyway, once you get toI would do this...
 &= 2 \\ \tan^3{x} + \tan{x} &= 2 \\ \tan^3{x} + \tan{x} - 2 &= 0 \\ X^3 + X - 2 &= 0 \textrm{ if we let }X = \tan{x} \\ \left(X - 1\right)\left(X^2 + X + 2\right) &= 0 \end{align*})
You should be able to solve this now. Also please advise if you only want real solutions or if you want complex solutions as well (they are, as the name suggests, more complex)...
tanx=1 gives x=pi/4 which cannot be a solution
I suggest the following sin2x/cos2x=sin2x sin2x=sin2xcos2x sin2x-sin2xcos2x =0 sin2x(1-cos2x)=0 sin2x=0 or 1-cos2x=0 cos2x=1
The first gives 2x=0 or pi or 2pi or 3pi So x= 0 or pi/2 or pi or3pi/2 cos2x=1 gives 2x= 0 or 2pi so x= 0 or pi which we already have.
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