We know that,
#color(violet)((1)sintheta=cos(pi/2-theta)#
#color(blue)((2)costheta=1=cos0=>theta=2kpi+-0=2kpi, kinZZ#
#color(red)((3)costheta=cosalpha=>theta=2kpi+-alpha,kinZZ#
Here,
#1+2cos^2(x+pi/6)=3color(violet)(sin(pi/3-x)...toApply(1)#
#=>1+2cos^2(x+pi/6)=3color(violet)(cos[pi/2-(pi/3-x)]#
#=>1+2cos^2(x+pi/6)=3cos[pi/2-pi/3+x]#
#=>1+2cos^2(x+pi/6)=3cos(pi/6+x)#
#=>2cos^2(x+pi/6)-3cos(x+pi/6)+1=0#
Let , #cos(x+pi/6)=m#
#:.2m^2-3m+1=0#
#=>2m^2-2m-m+1=0#
#=>2m(m-1)-1(m-1)=0#
#=>(m-1)(2m-1)=0#
#=>m-1=0 or 2m-1=0#
#=>m=1 or m=1/2,where, m=cos(x+pi/6)#
#=>cos(x+pi/6)=1 or cos(x+pi/6)=1/2#
#(i)cos(x+pi/6)=1#
#=>color(blue)(x+pi/6=2kpi, kinZZ...toApply(2)#
#=>x=2kpi-pi/6, kinZZ#
#(ii)cos(x+pi/6)=1/2=cos(pi/3)#
#=>color(red)(x+pi/6=2kpi+-pi/3.kinZZ...toApply(3)#
#=>x+pi/6=2kpi+pi/3 or x+pi/6=2kpi-pi/3 ,kinZZ#
#=>x=2kpi+pi/3-pi/6 orx=2kpi-pi/3-pi/6 ,kinZZ#
#=>x=2kpi+pi/6 ,or x=2kpi-pi/2 ,kinZZ#
Hence,
#x=2kpi-pi/6 orx=2kpi+pi/6 orx=2kpi-pi/2 ,kinZZ#
#i.e. x={2kpi+-pi/6,kinZZ}uu{2kpi-pi/2,kinZZ}#