The first step is to determine the domain of the functions involved. With the restriction #0<=x<=360# (degrees, I assume) we have exclude the points where #tan(x)# is not defined.

By definition, #tan(x)=sin(x)/cos(x)#

Therefore, we have to exclude points where #cos(x)=0#, that is the domain is described as

#x != 90# and #x != 270#

With the above restrictions in mind we can transform the equation as follows:

#3sin(x)+2sin(x)/cos(x) = 0#

The immediate temptation to reduce the above equation by #sin(x)# should be accompanied by checking if it can be equal to zero to avoid losing solutions.

Indeed, #sin(x)=0# can occur within a domain defined above.

It happens when

#x=0#, #x=180# and #x=360#.

So the three values above are solutions.

After specifying these three solutions we can reduce the equation by #sin(x)# getting

#3+2/cos(x)=0#,

from which follows

#cos(x)=-2/3#,

solutions of this are

#x = arccos(-2/3)# and #x = -arccos(-2/3)#

*CHECKING*

The first three solutions (#0#, #180# and #360#) cause both #sin(x)# and #tan(x)# to be equal to #0#, therefore the left side will be equal to #0#.

The next two values (#arccos(-2/3)# and #-arccos(-2/3)# result in the following:

#3sin(arccos(-2/3))+2sin(arccos(-2/3))/cos(arccos(-2/3)) =#

#= sin(arccos(-2/3))*[3+2/(-2/3)] = #

#=sin(arccos(-2/3))*[3-3] = 0#

#3sin(-arccos(-2/3))+2sin(-arccos(-2/3))/cos(-arccos(-2/3)) =#

#= sin(-arccos(-2/3))*[3+2/(-2/3)] = #

#=sin(-arccos(-2/3))*[3-3] = 0#

All is checked.