Bring the equation to standard form:
#(sin 2x + cos 2x)^2 - 1 = 0#
(sin 2x + cos 2x - 1)(sin 2x + cos 2x + 1) = 0
a. (sin 2x + cos 2x - 1) = 0
sin 2x + cos 2x = 1
Reminder: #sin 2x + cos 2x = sqrt2sin (2x + pi/4)#
#sqrt2sin (2x + pi/4) = 1 #
#sin (2x + pi/4) = 1/sqrt2 = sqrt2/2#
Trig table and unit circle -->
#(2x + pi/4) = pi/4 and (2x + pi/4) = (3pi)/4#
A. #2x + pi/4 = pi/4# --> 2x = 0 --> x = 0
B. #2x + pi/4 = (3pi)/4# --> #2x = pi/2# --> #x = pi/4#
b. sin 2x + cos 2x = - 1
#sqrt2.sin (2x + pi/4) = - 1#
#sin (2x + pi/4) = - 1/sqrt2 = - sqrt2/2#
Trig table and unit circle -->
#sin (2x + pi/4) = - (3pi)/4#, or #(5pi)/4# (co-terminal)
and #sin (2x + pi/4) = - pi/4#, or #(7pi)/4# (co-terminal)
C. #sin (2x + pi/4) = (5pi)/4# --> #2x = pi# --> #x = pi/2#
D. #(2x + pi/4) = (7pi)/4# --> #2x = (3pi)/2# --> #x = (3pi)/4#
Answers for #(0, 2pi)#:
#0, pi/4, pi/2, (3pi)/4#
