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.

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- Apr 29th 2010, 03:38 PMr8654help solving a trig equation
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. - Apr 29th 2010, 04:30 PMskeeter
$\displaystyle \tan(2x) - 2\sin{x} = 0$

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

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

$\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 ? - Apr 29th 2010, 04:55 PMr8654
yes thank you. The double angles always get me.