Question #c6dcd

1 Answer
Nghi N.
Mar 19, 2017

#x = 2kpi#
#x = pi/2 + 2kpi#.

Explanation:

Use trig identity:
#sin x + cos x = sqrt2sin (x + pi/4)#
In this case:
#sin x + cos x = sqrt2sin (x + pi/4) = 1#
#sin (x + pi/4) = 1/sqrt2 = sqrt2/2#
Two solutions:
#(x + pi/4) = pi/4 + 2kpi#, and #(x + pi/4) = (3pi)/4 + 2kpi#
a. #x + pi/4 = pi/4 + 2kpi#
#x = 0 + 2kpi#
b. #x + pi/4 = (3pi)/4 + 2kpi#
#x = (3pi)/4 - pi/4 = pi/2 + 2kpi.#

Check.
x = 0 --> sin o = 0 --> cos 0 = 1 --> 0 + 1 = 1 OK
#x = pi/2# --> #sin (pi/2) = 1# --> #cos (pi/2)# = 0 --> 1 + 0 = 1. OK

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