Whenever #cos(theta) = 1#,
we get #sin(theta) = +-sqrt(1-cos^2theta) = 0#
and #cos(theta)-sin(theta) = 1 - 0 = 1#.
#cos(theta) = 1# for #theta = 2npi# for all #n in ZZ#.
Whenever #sin(theta) = -1#,
we get #cos(theta) = +-sqrt(1-sin^2theta) = 0#
and #cos(theta)-sin(theta) = 0 - (-1) = 1#.
#sin(theta) = -1# for #theta = -pi/2+2npi# for all #n in ZZ#.
Putting two cases together, we have solutions when:
#theta = 2npi# for all #n in ZZ#
and when
#theta = -pi/2 + 2npi# for all #n in ZZ#
To make sure that these are the only solutions:
Starting with #cos(theta)-sin(theta)=1#, first add #sin(theta)# to both sides:
#cos(theta)=sin(theta)+1#
Then square both sides:
#cos^2(theta)=sin^2(theta)+2sin(theta)+1#
Then use #cos^2(theta)=1-sin^2(theta)# to get:
#1-sin^2(theta)=sin^2(theta)+2sin(theta)+1#
Add #sin^2(theta)-1# to both sides to get:
#0=2sin^2(theta)+2sin(theta)=2sin(theta)(sin(theta)+1)#
So either #sin(theta) = 0# or #sin(theta) = -1#
We have already accounted for #sin(theta) = -1# in our solutions.
What about #sin(theta) = 0#?
If this is so, then
#cos^2(theta) = 1 - sin^2(theta) = 1 - 0 = 1#
So #cos(theta) = +-sqrt(1) = +-1#
Only the case #cos(theta) = 1# satisfies #cos(theta)-sin(theta) = 1# and we have accounted for that case already too.
So we have found all the solutions.
