hello, can anyone help me solve the problem of Tan(2x)-2sin(x)=0
I know the double angles of tan is 2tan/1-tan^2. I get 2tan/1-tan^2 -2sin(x)=0
I have no idea what to do from there.
Thank you for any help given.
$\displaystyle \tan(2x) - 2\sin{x} = 0$
$\displaystyle \frac{\sin(2x)}{\cos(2x)} - 2\sin{x} = 0$
$\displaystyle \frac{2\sin{x}\cos{x}}{\cos(2x)} - 2\sin{x} = 0$
$\displaystyle 2\sin{x}\left[\frac{\cos{x}}{\cos(2x)} - 1\right] = 0
$
$\displaystyle 2\sin{x}\left[\frac{\cos{x} - \cos(2x)}{\cos(2x)}\right] = 0$
$\displaystyle 2\sin{x}\left[\frac{1 + \cos{x} - 2\cos^2{x}}{\cos(2x)}\right] = 0
$
$\displaystyle 2\sin{x}\left[\frac{(1+2\cos{x})(1-\cos{x})}{\cos(2x)}\right] = 0$
$\displaystyle \sin{x} = 0$
$\displaystyle \cos{x} = -\frac{1}{2}$
$\displaystyle \cos{x} = 1$
finish up ?