How to solve for x when restricted to interval [0,2pi] tan(x/2)=(2-sqrt(2))/2sinx?

Apr 3, 2018

$x = \frac{\pi}{4} \mathmr{and} x = \frac{7 \pi}{4}$

Explanation:

$\tan \left(\frac{x}{2}\right) = \frac{2 - \sqrt{2}}{2 \sin x}$
$\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} = \frac{2 - \sqrt{2}}{4 \sin \left(\frac{x}{2}\right) . \cos \left(\frac{x}{2}\right)}$
Simplify by $\cos \left(\frac{x}{2}\right)$ and cross-multiply -->
$4 {\sin}^{2} \left(\frac{x}{2}\right) = 2 - \sqrt{2}$
${\sin}^{2} \left(\frac{x}{2}\right) = \frac{2 - \sqrt{2}}{4}$
$\sin \left(\frac{x}{2}\right) = \pm \frac{\sqrt{2 - \sqrt{2}}}{2} = \pm 0.38$
Calculator and unit circle give 4 solutions-->
a. $\sin \left(\frac{x}{2}\right) = \left(\frac{\sqrt{2 - \sqrt{2}}}{2}\right) = 0.38$ -->
$\frac{x}{2} = \frac{\pi}{8} \implies x = \frac{\pi}{4} \in \left[0 , 2 \pi\right]$, and
$x = 2 \pi - \frac{\pi}{4} \implies x = \frac{7 \pi}{4} \in \left[0 , 2 \pi\right]$
b. $\sin \left(\frac{x}{2}\right) = - 0.38$
$x = - \frac{\pi}{4} \notin \left[0 , 2 \pi\right]$ and
$x = - \frac{7 \pi}{4} \notin \left[0 , 2 \pi\right]$
Hence,
$x = \frac{\pi}{4} \mathmr{and} x = \frac{7 \pi}{4}$