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Answer 1 · 2018-02-12T09:33:36

#x=0,2pi#

Your question is

#cos(x-pi/6) + cos(x+pi/6) = sqrt3# in the interval #[0,2pi]#.

We know from trig identities that

#cos(A+B) = cosAcosB-sinAsinB#
#cos(A-B) = cosAcosB+sinAsinB#

so that gives

#cos(x-pi/6) = cosxcos(pi/6)+sinxsin(pi/6)#
#cos(x+pi/6) = cosxcos(pi/6)-sinxsin(pi/6)#

therefore,

#cos(x-pi/6)+cos(x+pi/6)#
#=cosxcos(pi/6)+sinxsin(pi/6)+cosxcos(pi/6)-sinxsin(pi/6)#
#=2cosxcos(pi/6)#

So we now know we can simplify the equation to

#2cosxcos(pi/6) = sqrt3#

#cos(pi/6) = sqrt3/2#

so

#sqrt3cosx = sqrt3 -> cosx = 1#

We know that in the interval #[0,2pi]#, #cosx=1# when #x=0, 2pi#

Answer 2 · 2018-02-12T09:39:24

#"No soln. in "(0,2pi)#.

#cos(x-pi/6)+cos(x+pi/6)=sqrt3#

Using, #cosC+cosD=2cos((C+D)/2)cos((C-D)/2)#,

# 2cosxcos(-pi/6)=sqrt3#,

#:. 2*sqrt3/2*cosx=sqrt3#,

#:. cosx=1=cos0#.

Now, #cosx=cosy rArr x=2kpi+-y, k in ZZ#.

#:. cosx=cos0 rArr x=2kpi, k in ZZ, i.e., #

# x=0,+-2pi, +-4pi,...#

#:." The Soln. Set" sub (0,2pi)" is "phi#.