Which identity to change when solving an equation like this

**200001** · March 28th 2010, 01:11 AM

Hi
Where would you start?

I am trying to get to one trig function and used a few different identities but am going round in circles!
I though to express them all as cosine...then sine and I have just ended up going back on myself

**running-gag** · March 28th 2010, 05:09 AM

Hi

$\frac{2}{\cos(2x)}-\frac{\cos(2x)}{\sin(2x)} = \frac{\sin(2x)}{\cos(2x)}$

$\frac{2\:\sin(2x)-\cos^2(2x)-\sin^2(2x)}{\cos(2x)\:\sin(2x)} = 0$

$\frac{2\:\sin(2x)-1}{\cos(2x)\:\sin(2x)} = 0$

**200001** · March 28th 2010, 06:22 AM

Thanks
So that would leave me to solve arscin .05 and then divide all answers by two in the given range?

**running-gag** · March 28th 2010, 06:53 AM

Yes
It leads to $x=\frac{\pi}{12} [\pi]$ or $x=\frac{5\pi}{12} [\pi]$

**200001** · March 28th 2010, 07:54 AM

Brilliant
Thanks!

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