Question 760334
{{{sqrt(2)cos(x/2)-1=0}}} --> {{{cos(x/2)=1/sqrt(2)=sqrt(2)/2}}}
The angles that have a cosine value of {{{1/sqrt(2)=sqrt(2)/2}}} are in the first and fourth quadrant.
 
{{{x/2=pi/4}}} is a solution in the first quadrant
That gives us:
{{{x/2=pi/4}}} --> {{{highlight(x=pi/2)}}}
The next fist quadrant solution to the equation is
{{{x/2=pi/4+2pi=9pi/4}}} --> {{{x=9pi/2>2pi}}} which is not in the interval [{{{0}}},{{{2pi}}})
 
{{{x/2=-pi/4}}} is a fourth quadrant solution to {{{sqrt(2)cos(x/2)-1=0}}}
and so are all angles differing by a multiple of {{{2pi}}}
That makes {{{x/2=-pi/4+2pi=7pi/4}}} a solution to {{{sqrt(2)cos(x/2)-1=0}}}
However, {{{x/2=7pi/4}}} --> {{{x=7pi/2>2pi}}} which is not in the interval [{{{0}}},{{{2pi}}}).
 
SO the only solution in the interval [{{{0}}},{{{2pi}}}) is {{{highlight(x=pi/2)}}}.