# How do I solve cos(x/2)=sqrt(2)/2 with a domain of [-720º,360º)?

##### 1 Answer
Apr 12, 2018

$\textcolor{b l u e}{- {630}^{\circ} , - {90}^{\circ} , {90}^{\circ}}$

#### Explanation:

$\cos \left(\frac{x}{2}\right) = \frac{\sqrt{2}}{2}$

$\frac{x}{2} = \arccos \left(\cos \left(\frac{x}{2}\right)\right) = \arccos \left(\frac{\sqrt{2}}{2}\right)$

$\implies \frac{x}{2} = {45}^{\circ} , {315}^{\circ}$

$\frac{x}{2} = {45}^{\circ} + n {360}^{\circ} \implies x = {90}^{\circ} + n {720}^{\circ}$

$\frac{x}{2} = {315}^{\circ} + n {360}^{\circ} \implies x = {630}^{\circ} + n {720}^{\circ}$

For:

$n \in \mathbb{Z}$

We now use:

$x = {90}^{\circ} + n {720}^{\circ} \mathmr{and} x = {630}^{\circ} + n {720}^{\circ}$

with integer n values to find angles in the interval:

$\left[- {720}^{\circ} , {360}^{\circ}\right)$

$n = 0 \to {90}^{\circ} , {630}^{\circ}$

$n = - 1 \to - {630}^{\circ} , - {90}^{\circ}$
$n = - 2 \to - {1350}^{\circ} , - {810}^{\circ}$

So only the values:

$\textcolor{b l u e}{- {630}^{\circ} , - {90}^{\circ} , {90}^{\circ}}$