Question 609385
{{{cos x-sin(x/2) = 0}}}
Transform Cos into Sin using Double Angle Identity: {{{cos(2u) = 1-2sin^2(u)}}}
Let {{{u = x/2}}}, so {{{cos(x) = 1-2sin^2(x/2)}}}
{{{1-2sin^2(x/2) - sin(x/2) = 0}}}
{{{-2sin^2(x/2)-sin(x/2)+1 = 0}}}
{{{2sin^2(x/2)+sin(x/2)-1 = 0}}}
Factor: {{{(sin(x/2)+1)(2sin(x/2)-1) = 0}}}
Split into two equations:
{{{sin(x/2)+1 = 0}}} and {{{2sin(x/2)-1 = 0}}}
For {{{sin(x/2)+1 = 0}}}:
{{{sin(x/2) = -1}}}
Take the inverse Sin:
{{{x/2 = 2n*pi-pi/2}}}, where n is an integer
{{{x = 4n*pi-pi}}}, so no value of n is within the restriction 0 to {{{2pi}}}
not even n=0, since {{{x = -pi}}}
NOTE: x can't equal positive {{{pi}}} since {{{sin(pi/2) <> -1}}}
and the given equation: {{{sin(x/2)-cos(x)=0}}} would not be true!

For {{{2sin(x/2)-1 = 0}}}:
{{{2sin(x/2) = 1}}}
{{{sin(x/2) = 1/2}}}
Take the inverse Sin:
{{{x/2 = pi/6+2n*pi}}}
{{{x = pi/3+4n*pi}}}, n must = 0, else not within restriction, so {{{x = pi/3}}}
OR
{{{x/2 = 5pi/6+2n*pi}}}
{{{x = 5pi/3+4n*pi}}}, n must = 0, else not within restriction, so {{{x = 5pi/3}}}

So all values of x within restriction 0 to {{{2pi}}} are:
({{{pi/3}}}, {{{5pi/3}}})