Originally Posted by

**topsquark** You can't solve this. It isn't an equation. You could simplify it though...

$\displaystyle ln(x) - ln(x - 1) + 1 = ln(x) - ln(x - 1) + ln(e) = ln \left ( \frac{ex}{x - 1} \right )$

$\displaystyle sin(2x) = 2sin(x)cos(x)$

so...

$\displaystyle sin(2x) - sin(x) = 0$

$\displaystyle 2sin(x)cos(x) - sin(x) = 0$

$\displaystyle sin(x) ( 2cos(x) - 1 ) = 0$

So either

$\displaystyle sin(x) = 0 \implies x = 0, ~ \pi$

or

$\displaystyle 2cos(x) - 1 = 0 \implies cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, ~ \frac{5\pi}{3}$

-Dan