Given: #sqrt(1+cos(x))-sqrt(1-sin(x))=1#
Square both sides:
#1 + cos(x) -2sqrt(1+cos(x))sqrt(1-sin(x)) + 1 - sin(x) = 1#
#cos(x) -sin(x) + 1 = 2sqrt(1+cos(x))sqrt(1-sin(x))#
Square both sides:
#cos^2(x) - sin(x)cos(x)+cos(x)-sin(x)cos(x)+sin^2(x)-sin(x)+cos(x)-sin(x)+1 = 4(1+cos(x))(1-sin(x))#
# 2cos(x)- 2sin(x)cos(x)-2sin(x)+2 = 4(1+cos(x))(1-sin(x))#
# cos(x)- sin(x)cos(x)-sin(x)+1 = 2(1+cos(x)-sin(x) - sin(x)cos(x))#
# cos(x)- sin(x)cos(x)-sin(x)+1 = 2+2cos(x)-2sin(x) - 2sin(x)cos(x)#
# cos(x)- sin(x)cos(x)-sin(x)+1 = 0#
Factor:
#(1 + cos(x))(1 - sin(x)) = 0#
#cos(x) = -1 and sin(x) = 1#
check #cos(x) = -1# then #sin(x) = 0#:
#sqrt(1-1)-sqrt(1-0)=1#
#-1 = 1#
This does not check; discard the root #cos(x) = -1#
check #sin(x) = 1#, then #cos(x) = 0#:
#sqrt(1-0)-sqrt(1-1)=1#
#1 = 1#
This checks
Find repetitive values of x where #sin(x) =1#
#x = sin^-1(1)#
#x = pi/2#
This repeats at integer multiples of #2pi#:
#x = pi/2 + 2pin; n in ZZ#
