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}