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QUESTION
How do you solve ##sinx+cosx=1##?
##x = 2npi and x = (4n+1)pi/2, n = 0, +-1, +-2, +-3. ...##
The given equation is equivalent to
##1/sqrt 2 sin x+ 1/sqrt 2 cosx=1/sqrt 2##.
This can be written as
##cos (x-pi/4) = cos (pi/4)##
The general solution of this equation ls
##x-.pi/4=2npi+-pi/4, n = 0, +-1, +-2, ...##,
So, ##x = 2npi and x = (4n+1)pi/2, n = 0, +-1, +-2, +-3. ...##
