Question 138440
I suspect you are frustrated because you are trying to make this too difficult.  If you are using the double angle formulas you are barking up the wrong tree.


Remember that {{{sin(alpha)/cos(alpha)=tan(alpha)}}}


Let {{{alpha=2x}}}


Then {{{sqrt(3)*sin(2x)=cos(2x)}}} => {{{sqrt(3)*sin(alpha)=cos(alpha)}}} 


Now with a bit of algebra we can write: {{{sin(alpha)/cos(alpha)=1/sqrt(3)}}} which is to say {{{tan(alpha)=1/sqrt(3)}}} => {{{alpha=arctan(1/sqrt(3))}}} => {{{alpha=pi/6}}}


But {{{alpha=2x}}} , so {{{2x=pi/6}}} => {{{x=pi/12}}}.  The periodicity of {{{tan(x)}}} is {{{pi}}}, so the complete solution is {{{x=pi/12+k*pi}}} where k is any integer.