We start as if this wasn't a trigonometric equation:

#5sinx+2=sinx# For now we'll 'pretend' that its domain is #D_0=\mathbb{R}# (all #x#s that make sense) and at the end we'll adjust the solutions to its actual domain #D=[0,2pi]#.

#5sinx+2=sinx#

#5sinx-sinx=-2#

#4sinx=-2#

#sinx=-2/4#

#sinx=-1/2#

Now, here we have at least two ways to go, I personally would use the fact that sine is an odd function:

#-sinx=1/2#

#sin(-x)=1/2#

#-x=pi/6+2kpi or -x=pi-pi/6+2kpi, k in mathbbZ#

#x=-pi/6+2kpi or x=-(5pi)/6+2kpi, k in mathbbZ#

(Note: during the last step we multiplied both equations by #(-1)# and yet #+2kpi# stayed #+2kpi# - didn't change to #-2kpi#. The reason for that it that here it doesn't really matter because #k# goes trough the set of all integers - both positive and negative. I'm just used to writing #+somepi#)

Now we can adjust our solution to the domain #D=[0,2pi]#. Concluding:

#x in {(7pi)/6,(11pi)/6}#