Which identity to change when solving an equation like this

• Mar 28th 2010, 01:11 AM
200001
Which identity to change when solving an equation like this
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 myselfhttp://img532.imageshack.us/img532/3382/trig22.jpg
• Mar 28th 2010, 05:09 AM
running-gag
Hi

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

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

$\displaystyle \frac{2\:\sin(2x)-1}{\cos(2x)\:\sin(2x)} = 0$
• Mar 28th 2010, 06:22 AM
200001
Thanks
So that would leave me to solve arscin .05 and then divide all answers by two in the given range?
• Mar 28th 2010, 06:53 AM
running-gag
Yes
It leads to $\displaystyle x=\frac{\pi}{12} [\pi]$ or $\displaystyle x=\frac{5\pi}{12} [\pi]$
• Mar 28th 2010, 07:54 AM
200001
Brilliant
Thanks!