We can firstly factorise our original equation to make it become:
#sintheta(1+2costheta)=0#
Therefore, if we set the two parts to zero separately that will make the equation work. For example, if #sintheta=0#, then #0*#(whatever the other part becomes) will indeed #=0#.
So, separately we will make the two parts #=0#. Starting with the left part, if #sintheta=0#, #theta=0^o, 180^o,360^o#... For the remainder of this question I will consider only the domain #0^o<=theta<=360^o#.
Then, we make the other part #=0#. This means that #1+2costheta=0#, by rearranging this becomes #costheta=-1/2#. By solving with a calculator, our principal value is #120^o#. #240^o# is also a solution since if we consider how #costheta# reflects then we can obtain the above solution by doing #360-120#.
Our final solutions for #theta# are #0^o, 120^o, 180^o, 240^o and 360^o#, within the restricted domain #0^o<=theta<=360^o#. You can test all of these by substituting them into the original equation and they will indeed produce zero.
