Question #92a25

2 Answers
P dilip_k
Apr 3, 2017

#x=0,pi/2,2pi#

Explanation:

#1-sinx=cosx#

#=>cosx+sinx=1#

#=>1/sqrt2cosx+1/sqrt2sinx=1/sqrt2#

#=>cos(pi/4)cosx+sin(pi/4)sinx=cos(pi/4)#

#=>cos(x-pi/4)=cos(pi/4)=cos(-pi/4)=cos(2pi-pi/4)#

So #x-pi/4=pi/4#

#=>x=pi/2#

when

#x-pi/4=-pi/4#

#x=0#

Again

#x-pi/4=2pi-pi/4#

#=>x=2pi#

Nghi N
Apr 4, 2017

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

Explanation:

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

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