Question #483b0

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Answer 1 · 2016-11-06T07:41:46

#x=pi+2kpi or x=pi/4+kpi#

Since #cos^2(x/2)=(1+cosx)/2 and tanx=sinx/cosx#,

by substituting you have:

#cancel2(1+cosx)/cancel2=sinx/cosx+sinx#

#cosx+cos^2x=sinx+sinxcosx and cosx!=0#

#cosx+cos^2x-sinx-sinxcosx=0 and cosx!=0#

#cosx(1+cosx)-sinx(1+cosx)=0 and cosx!=0 #

#(1+cosx)(cosx-sinx)=0#

#1+cosx=0 or cosx-sinx=0#

#cosx=-1 or cosx=sinx#

#x=pi+2kpi or x=pi/4+kpi#