How to solve #1+2cos^2(x+pi/6)=3sin(pi/3-x)#?

#1 + cos^2 (x + pi/6) = 3sin (pi/3 - x) (1)#
Note.
#3sin (pi/3 - x) = 3cos (pi/2 - (pi/3 - x) = 3cos (x + pi/6)#.
Call #(x + pi/6) = t# . The equation (1) becomes:
#2cos^2 t - 3cos t + 1 = 0#.
Solve this quadratic equation for cos t.
Since a + b + c = 0, use shortcut. The 2 real roots are:
#cos t = 1# and #cos t = c/a = 1/2#

a. cos t = cos (x + pi/6) = 1
#x + pi/6 = 2pi#
#x = 2pi - pi/6 = (5pi)/6 + 2kpi#
b. #cos (x + pi/6) = 1/2#
#x + pi/6 = +- pi/3#
1. #x + pi/6 = pi/3# --> #x = pi/3 - pi/6 = pi/6 + 2kpi#
2. #x + pi/6 = (5pi)/3# (co-terminal to #x = - pi/3#)
#x = (5pi)/3 - pi/6 = (9pi)/6 = (3pi)/2 + 2kpi#

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}#