Apply the trig identity:
#sin a + cos a = sqrt2 cos (a - pi/4)# (found in any trig book)
#sin x + cos x = sqrt2cos (x - pi/4) = -1#
Trig table, and unit circle give -->
#cos (x - pi/4) = -1/sqrt2 = - sqrt2/2# --> #x = +- (3pi)/4#
Two solutions:
a. #(x - pi/4) = (3pi)/4# --> #x = (3pi)/4 + pi/4 = pi#
b. Since the arc #-(3pi)/4# --> #arc (5pi)/4# (co-terminal arcs), therefor:
#cos (x - pi/4) = (5pi)/4# --> #x = ((5pi)/4 + pi/4) = (6pi)/4 = (3pi)/2#
Answers for #(0, 2pi)#
#pi, and (3pi)/2#
Check:
#x = pi# --> sin x = 0 --> cos x = -1 --> -1 = -1 .OK
#x = (3pi)/2# --> sin x = -1 --> cos x = 0 --> -1 = -1 .OK
