Question 1009675
THE EXPECTED SOLUTION:
I do not know how you are expected to solve this.
No idea what is meant by "Round terms to three decimal places where appropriate", either.
Are you expected to round {{{sqrt(3)}}} or {{{2sqrt(3)}}} to 3-decimal places? Are you expected to find the solution values for {{{theta}}} in radians rounded to 3-decimal places? In degrees?
 
MY SOLUTION:
I will temporarily rename variables, so as not to have to keep writing {{{sin(theta)}}} and {{{cos(theta)}}} so many times.
{{{x=cos(theta)}}} and {{{y=sin(theta)}}} .
With those changes, the equation looks like
{{{4xy+2sqrt(3)y-2sqrt(3)x-3=0}}}
Now I rearrange and factor by parts:
{{{(4xy-2sqrt(3)x)+(2sqrt(3)y-3)=0}}}
{{{2x(2y-sqrt(3))+sqrt(3)(2y-sqrt(3))=0}}}
{{{(2x+sqrt(3))(2y-sqrt(3))=0}}}-->{{{system(2x+sqrt(3)=0,"or",2y-sqrt(3)=0)}}}-->{{{system(x=-sqrt(3)/2,"or",y=sqrt(3)/2)}}} .
So, going back to {{{theta}}}
{{{system(cos(theta)=-sqrt(3)/2,"or",sin(theta)=sqrt(3)/2)}}}
In the interval {{{"[ 0 ,"}}}{{{2pi}}}{{{"]"}}} , the solutions for{{{theta}}} (in radians) are
{{{cos(theta)=-sqrt(3)/2}}}-->{{{theta=pi +- pi/6}}} (rounds to 2.618 and 3.665).
{{{sin(theta)=-sqrt(3)/2}}}-->{{{theta=pi/2 +- pi/6}}} (rounds to 1.047 and 2.094).