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How to find $x$ in some trigonometric equations

How to solve these trigonometric equations?

$$\tan2x-\sin4x = 0$$

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marked as duplicate by Américo Tavares, J. M., David Mitra, Cocopuffs, Jennifer Dylan Aug 17 '12 at 16:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Why did you post the same equation of your last question? –  Américo Tavares Aug 17 '12 at 16:07
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My hint and answer to the last question (as linked by Américo Tavares) give a method of solving this. –  Mark Bennet Aug 17 '12 at 16:09

2 Answers 2

HINT: $\sin4x=2\sin2x\cos2x$, and $\tan2x=\dfrac{\sin2x}{\cos2x}$. Now let $a=\sin2x$ and $b=\cos2x$, write your equation in terms of $a$ and $b$, and see what it tells you about $a$ and $b$.

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Since this is a homework, some intermediate steps are omitted and left for you to work out.

The identities you need are$^{\dagger}$ $$ \sin(4x) = \color{red}{2} \sin(2x) \cos(2x)\\ \tan(2x) = \frac{\sin(2x)}{\cos(2x)} $$ Substitute both in $$ \tan(2x)-\sin(4x) = 0$$ to get $$ (1 - 2 \cos^2(2x))\sin(2x) = 0 $$ which you should be able to factor into $3$ cases. Solve each case for $x.$


$^{\dagger}$ Fixed error thanks to Thomas Andrews and David Mitra.

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can u show me the step to get $$ (1 - 2 \cos^2(2x))\sin(2x) = 0 $$ ? I really don't know and what should i do next? –  dramasea Aug 17 '12 at 16:10
    
@dramasea If substitute as Andrew said in his answer, you'll get $\cfrac {\sin 2x}{\cos 2x} = 2 \sin 2x \cos 2x.$ Now multiply both sides by $\cos 2x$ and factor. –  user2468 Aug 17 '12 at 16:12
    
As for the steps after, you have 3 cases: $\sin(2x) = 0,$ and two cases for $\cos^2(2x) = 1/2.$ Try to work them out. –  user2468 Aug 17 '12 at 16:13
    
I was wondering - did you choose your username after the character from the TV series Scrubs? –  Martin Sleziak Aug 17 '12 at 18:53

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