Use the modified Pythagorean identity:
#sin^2x+cos^2x=1#
#=>sin^2x=1-cos^2x#
First, square both sides of the equation:
#sinx = 1 - cosx#
#sin^2x=(1-cosx)^2#
#sin^2x=1-2cosx+cos^2x#
#0=1-2cosx+cos^2x-sin^2x#
#0=1-2cosx+cos^2x-(1-cos^2x)#
#0=1-2cosx+cos^2x-1+cos^2x#
#0=1-1-2cosx+2cos^2x#
#0=2cos^2x-2cosx#
#0=cos^2x-cosx#
#0=cosx(cosx-1)#
#cosx=0, cosx=1#
The solutions for #cosx=0# are #pi/2# and #(3pi)/2#, and the solution for #cosx=1# is #0#.
Now, plug all these values back into the original equation to make sure they hold true:
#sin(pi/2) stackrel(?)= 1-cos(pi/2)#
#1stackrel(?)=1-0#
#1=1#
The solution #pi/2# works.
#sin((3pi)/2)stackrel(?)=1-cos((3pi)/2)#
#-1stackrel(?)=1-0#
#-1!=1#
The solution #(3pi)/2# doesn't work.
#sin(0)stackrel(?)=1-cos(0)#
#0stackrel(?)=1-1#
#0=0#
The solution #0# works.
The final solutions are #pi/2# and #0#. Now, we have to add #2pik# to each of them, to represent that the solution would still be valid after any full rotation (#k# represents any integer). The full answer is now:
#x=0+2pik,pi/2+2pik#
