# How do you find all solutions of tan(x/2)-sinx=0 in the interval [0,2pi)?

Nov 14, 2017

$0 , \frac{\pi}{2} , \pi , \frac{3 \pi}{2}$

#### Explanation:

$\tan \left(\frac{x}{2}\right) = \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} = \frac{\sqrt{\frac{1}{2} \left(1 - \cos x\right)}}{\sqrt{\frac{1}{2} \left(1 + \cos x\right)}}$

$\frac{\sqrt{\frac{1}{2} \left(1 - \cos x\right)}}{\sqrt{\frac{1}{2} \left(1 + \cos x\right)}} - \sin x = 0$

${\left(\frac{\sqrt{\frac{1}{2} \left(1 - \cos x\right)}}{\sqrt{\frac{1}{2} \left(1 + \cos x\right)}}\right)}^{2} = {\sin}^{2} x$

$\frac{1 - \cos x}{1 + \cos x} = {\sin}^{2} x$

$1 - \cos x = {\sin}^{2} x + {\sin}^{2} x \cos x$

$1 - \cos x = 1 - {\cos}^{2} x + {\sin}^{2} x \cos x$

$- \cos x = - {\cos}^{2} x + {\sin}^{2} x \cos x$

${\cos}^{2} x - {\sin}^{2} x \cos x - \cos x = 0$

$\cos x \left(1 - {\sin}^{2} x - 1\right) = 0$

$\cos x \left(- {\sin}^{2} x\right) = 0$

$\cos x = 0 \implies x = \frac{\pi}{2} , \frac{3 \pi}{2}$

${\sin}^{2} x = 0 \implies x = 0 , \pi$