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How to solve these trigonometric equations?

$$\tan2x-\sin4x = 0$$

and

$$\tan2x = \sin x$$

I can't do this, please help me! I did this:

\begin{align} \tan2x &= 2\sin x\\ \\ \frac{\sin2x}{\cos2x} &= \tan x \end{align}

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    $0, (1/2)\pi, (1/8)\pi, (3/8)\pi, -(1/8)\pi, -(3/8)\pi;0, \pi, (2/3)\pi, -(2/3)\pi$.2012-08-17
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    how to do that bro2012-08-17
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    $\sin {2x} = \cfrac {2\tan x} {1+\tan^2 x}$2012-08-17
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    @MarkBennet That is $\tan 2x$, no?2012-08-17
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    @MichaelBoratko I believe he is using the [Weierstrass substitution](http://en.wikipedia.org/wiki/Weierstrass_substitution).2012-08-17
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    @DanLitt Indeed - I read the denominator with a $-$ instead of the $+$ which is there. Apologies!2012-08-17
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    @MichaelBoratko An easy mistake to make!2012-08-17
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    I have put a solution (incomplete, but the legwork is straightforward) to show how to use the Weierstrass formula to reduce the first equation given to simple cases.2012-08-17

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