Question #10e97

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Answer 1 · 2017-05-13T14:07:37

The solution is #S={3/4pi+2kpi,7/4pi+2kpi}#, #k in ZZ#

We need

#sin^2x+cos^2x=1#

#sin(2x)=2sinxcosx#

#(a-b)^2=a^2-2ab-b^2#

Therefore,

#(sinx-cosx)=sqrt2#

Squaring both sides

#(sinx-cosx)^2=(sqrt2)^2#

#sin^2x-2sinxcosx+cos^2x=2#

#1-sin(2x)=2#

#sin2x=-1#

#2x=3/2pi# or #2x=7/2pi#

#x=3/4pi+2kpi#, #k in ZZ#

#x=7/4pi+2kpi#, #k inZZ#