# How do you find all solutions of sin(x/2)+cosx-1=0 in the interval [0,2pi)?

Jul 14, 2017

$x = 0 , \frac{\pi}{3} , 5 \frac{\pi}{3.}$

#### Explanation:

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

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

$\therefore \sin \left(\frac{x}{2}\right) - 2 {\sin}^{2} \left(\frac{x}{2}\right) = 0.$

$\therefore \sin \left(\frac{x}{2}\right) \left\{1 - 2 \sin \left(\frac{x}{2}\right)\right\} = 0.$

$\therefore \sin \left(\frac{x}{2}\right) = 0 , \mathmr{and} , \sin \left(\frac{x}{2}\right) = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right) .$

$\text{But, } x \in \left[0 , 2 \pi\right) \therefore 0 \le x < 2 \pi \therefore 0 \le \frac{x}{2} < \pi .$

$\therefore \sin \left(\frac{x}{2}\right) = 0 \Rightarrow \frac{x}{2} = 0 \therefore x = 0.$

Similarly, $\sin \left(\frac{x}{2}\right) = \sin \left(\frac{\pi}{6}\right) \Rightarrow \frac{x}{2} = \frac{\pi}{6} , \pi - \frac{\pi}{6} = 5 \frac{\pi}{6.}$

$\therefore x = \frac{\pi}{3} , 5 \frac{\pi}{3.}$

Altogether, $x = 0 , \frac{\pi}{3} , 5 \frac{\pi}{3.}$