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

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Answer 1 · 2018-04-12T11:40:53

$- 630^{\circ}, - 90^{\circ}, 90^{\circ}$ $color(blue)(-630^@,-90^@,90^@)$

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

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

$\Rightarrow \frac{x}{2} = 45^{\circ}, 315^{\circ}$ $=>x/2=45^@, 315^@$

$\frac{x}{2} = 45^{\circ} + n 360^{\circ} \Rightarrow x = 90^{\circ} + n 720^{\circ}$ $x/2=45^@+n360^@=>x=90^@+n720^@$

$\frac{x}{2} = 315^{\circ} + n 360^{\circ} \Rightarrow x = 630^{\circ} + n 720^{\circ}$ $x/2=315^@+n360^@=>x=630^@+n720^@$

For:

$n in ZZ$

We now use:

$x=90^@+n720^@ and x=630^@+n720^@$

with integer n values to find angles in the interval:

$[-720^@,360^@)$

$n=0-> 90^@, 630^@$

$n=-1 -> -630^@, -90^@$
$n=-2 -> -1350^@, -810^@$

So only the values:

$color(blue)(-630^@,-90^@,90^@)$