First use the double angle formula you suggested and the identity #cottheta = 1/tantheta#
#3((2tanx)/(1 - tan^2x)) - 3/tanx = 0#
#(6(sinx/cosx))/(1 - sin^2x/cos^2x) - 3/(sinx/cosx) = 0#
#((6sinx)/cosx)/((cos^2x - sin^2x)/cos^2x) - (3cosx)/sinx = 0#
#(6sinxcosx)/(cos^2x - sin^2x) - (3cosx)/sinx = 0#
#(6sinxcosx)/(1 - sin^2x - sin^2x) - (3cosx)/sinx = 0#
#(6sinxcosx)/(1 - 2sin^2x) - (3cosx)/sinx = 0#
#6sin^2xcosx - 3cosx + 6cosxsin^2x = 0#
#cosx(6sin^2x - 3 + 6sin^2x) = 0#
#cosx(12sin^2x - 3) = 0#
#cosx(3(4sin^2x - 1)) = 0#
#cosx= 0" AND "sinx = +-1/2#
#x = pi/2, (3pi)/2, pi/6, (5pi)/6, (7pi)/6, (11pi)/6#
I have checked these solutions and they all seem to work.
Hopefully this helps!
