#2tan(x)cos(3x)=-sqrt(3)tan(x)#
Move everything to one side
#2tan(x)cos(3x)+sqrt(3)tan(x)=0#
Factor #tan(x)# out
#tan(x)(2cos(3x)+sqrt(3))=0#
By zero product property, we have
#tan(x)=0# or #2cos(3x)+sqrt(3)=0#
#tan(x)=0# has solutions #x=kpi#, where #k# is an integer.
Since #x in [0,2pi)#, only 0 and #pi# are solutions.
Moving on with the other term
#2cos(3x)+sqrt(3)=0#
#cos(3x)+sqrt(3)/2=0#
#cos(3x)=-sqrt(3)/2#
Here a graph of #cos(x)# would be helpful - it gives a little suggestion (click and drag for (x,y) values and special points)
graph{(y-cos(x))(y+sqrt(3)/2)=0 [-1,7,-2,2]}
We need to recall, that #sqrt(3)/2=cos(pi/6)#
From the graph we see, that the solution is #3x=2kpi+pi pm pi/6# or #x=((2k+1)/3pm1/18)pi#
Since #x in [0,2pi)#, only solutions are
#(1/3-1/18)pi,(1/3+1/18)pi,(3/3-1/18)pi,(3/3+1/18)pi,(5/3-1/18)pi,(5/3+1/18)pi#
or simplified
#5/18pi,7/18pi,17/18pi,19/18pi,29/18pi,31/18pi#
