Question 372386
From double -angle identity, {{{cosx = 1 - 2(sin(x/2))^2}}}.  By direct substitution, 
{{{sin(x/2)+cosx-1= sin(x/2) + 1 - 2(sin(x/2))^2 - 1 = sin(x/2) - 2(sin(x/2))^2 = sin(x/2)*(1 - 2sin(x/2)) = 0}}}. Hence 
{{{sin(x/2) = 0}}}, or {{{1 - 2sin(x/2) = 0}}}.  (The 2nd is the same as {{{sin(x/2) = 1/2}}}.)
From the 1st equation we get x = 0 degrees, or 360 degrees.  From the second, we get x = 60 degrees, or 300 degrees.





****The solution given by the author below says that I have given ONLY TWO solutions, when I have given FOUR, namely 0, 60, 300, or 360 degrees.  If he had read my ENTIRE solution he should HAVE REALIZED that I was going for solutions between 0 and 360 degrees, inclusive.  I have left it to asker to just find the OTHER INFINITE solutions for his consideration.****