Question 630941
{{{sin(x/2)+cos x = 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)}}}
{{{sin(x/2)+1-2sin^2(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 = pi/2}}}
{{{x = pi}}}

For {{{2sin(x/2)+1 = 0}}}:
{{{2sin(x/2) = -1}}}
{{{sin(x/2) = -1/2}}}
Take the inverse Sin:
{{{x/2 = 7pi/6}}}
{{{x = 7pi/3}}}, but this is not within the restriction 0 to {{{2pi}}}
OR
{{{x/2 = 11pi/6}}}
{{{x = 11pi/3}}}, but this is not within the restriction 0 to {{{2pi}}}

So the only value of x within the restriction 0 to {{{2pi}}} is {{{pi}}}